Topology without Tears - 2.2 Definitions and Propositions

Proposition:

A subset \(S\) of \(\mathbb{R}\) is open if and only if it is a union of open intervals.

Proof:

• We shall first prove that a union of open intervals is an open set in \(\mathbb{R}\). Let \(S = \bigcup_{\alpha \in \Gamma} (a_{\alpha},b_{\alpha})\) where \((a_{\alpha},b_{\alpha})\) is an open interval in \(\mathbb{R}\). We proved in the previous section that any open interval is an open set. From the definition of a topology, we have any arbitrary union of open sets to be an open set. Hence, \(S = \bigcup_{\alpha \in \Gamma} (a_{\alpha},b_{\alpha})\) is an open set.
• Now we shall now prove that any open set in \(\mathbb{R}\) is a union of open intervals. From the definition of an open set in \(\mathbb{R}\), we have that for every \(x \in S\), we can find an open interval \((a_x,b_x)\) containing \(x\) such that \((a_x,b_x) \subseteq S\). Now consider \(\cup_{x \in S} (a_x,b_x)\). We shall now prove that \(S = \cup_{x \in S} (a_x,b_x)\).

• We shall prove first that \(S \subseteq \cup_{x \in S} (a_x,b_x)\). Consider any \(y \in S\). We hence have \(y \in (a_y,b_y)\). Hence, \(y \in \cup_{x \in S} (a_x,b_x)\). Hence, \(S \subseteq \cup_{x \in S} (a_x,b_x)\).
• We shall prove next that \(S \supseteq \cup_{x \in S} (a_x,b_x)\). Consider any \(y \in \cup_{x \in S} (a_x,b_x)\). This means that \(y \in (a_x,b_x)\) for some \(x \in S\). But \((a_x,b_x) \subseteq S\). Hence, we get \(y \in S\). Hence, \(S \supseteq \cup_{x \in S} (a_x,b_x)\).

This gives us that a subset \(S\) of \(\mathbb{R}\) is open if and only if it is a union of open intervals.

Definition:

Let \((X, \tau)\) be a topological space. A collection \(\mathcal{B}\) of open subsets of \(X\) is said to be a basis for the topology \(\tau\) if every open set is a union of members of \(\mathcal{B}\).

Proposition:

Let \((X,\tau)\) be a discrete space and \(\mathcal{B}\) the family of all singleton subsets of \(X\); that is, \(\mathcal{B} = \{\{x\}: x \in X\}\). Then \(\mathcal{B}\) is a basis for \(\tau\).

Proof:

The discrete topology contains all the subsets of \(X\). Note that any subset, \(A \subseteq X\), can be written as \(A = \cup_{x \in A} \{x\}\). Hence, the singleton set generate the topology.

Proposition:

There are many different bases for the same topology. If \(\mathcal{B}\) is a basis for a topology \(\tau\) on a set \(X\) and \(\mathcal{B}_1\) is a collection of subsets of \(X\) such that \(\mathcal{B} \subseteq \mathcal{B}_1 \subseteq \tau\), then \(\mathcal{B}_1\) is also a basis for \(\tau\).

Proof:

The topology generated by \(\mathcal{B}\) is \(\tau\). Let \(\tau(\mathcal{B}_1)\) be the topology generated by \(\mathcal{B}_1\). We need to show that \(\tau(\mathcal{B}_1) = \tau\).

• We shall first show that \(\tau(\mathcal{B}_1) \supseteq \tau\). We are given that \(\mathcal{B} \subseteq \mathcal{B}_1\) and \(\tau(\mathcal{B}) = \tau\). This means that any element in \(\tau\) can be written as \(\bigcup_{A_{\alpha} \in \mathcal{B}} A_{\alpha}\). Since any element in \(\mathcal{B}\) is also an element of \(\mathcal{B}_1\), we have that any element in \(\tau\) can be written as \(\bigcup_{A_{\alpha} \in \mathcal{B}_1} A_{\alpha}\). Hence, we have \(\tau(\mathcal{B}_1) \supseteq \tau\).
• Now we shall show that \(\tau(\mathcal{B}_1) \subseteq \tau\). We are given that \(\mathcal{B}_1 \subseteq \tau\). Let us look at the topology generated by \(\mathcal{B}_1\). Any element in the \(\tau(\mathcal{B}_1)\) is an arbitrary union of elements in \(\mathcal{B}_1\). Consider any arbitrary union of elements in \(\mathcal{B}_1\) i.e. \(\bigcup_{A_{\alpha} \in \mathcal{B}_1} A_{\alpha}\). Since \(\mathcal{B}_1 \subseteq \tau\), we have that \(A_{\alpha} \in \tau\) for all \(A_{\alpha} \in \mathcal{B}_1\). Since \(\tau\) is a topology, it is closed under arbitrary unions and hence we get \(\bigcup_{A_{\alpha} \in \mathcal{B}_1} A_{\alpha} \in \tau\). Hence any element in \(\tau(\mathcal{B}_1)\) is also an element in \(\tau\). Hence, we get \(\tau(\mathcal{B}_1) \subseteq \tau\). 

Hence, we get that \(\mathcal{B}_1\) is also a basis for \(\tau\).

Proposition:

Let \(X\) be a non-empty set and let \(\mathcal{B}\) be a collection of subsets of \(X\). Then \(\mathcal{B}\) is a basis for a topology on \(X\) if and only if \(\mathcal{B}\) has the following properties:

1. \(X = \bigcup_{B \in \mathcal{B}} B\)
2. For any \(B_1,B_2 \in \mathcal{B}\), then the set \(B_1 \cap B_2\) is a union of members of \(B\).

Proof:

If \(\mathcal{B}\) is a basis for a topology \(\tau\), then any element in \(\tau\) can be written as an arbitrary union of elements in \(\mathcal{B}\). Since, \(X \in \tau\), we have that \(X = \bigcup_{\alpha \in \Gamma} B_{\alpha}\) where \(B_{\alpha} \in \mathcal{B}\). But \(\bigcup_{\alpha \in \Gamma} B_{\alpha} \subseteq \bigcup_{B \in \mathcal{B}} B\). Hence, we get \(X \subseteq \bigcup_{B \in \mathcal{B}} B\). Further, for any \(B \in \mathcal{B}\), we have \(B \subseteq X\). Hence, we get \(\bigcup_{B \in \mathcal{B}} \subseteq X\). Hence, we have \(X = \bigcup_{B \in \mathcal{B}} B\). Also, for any \(B_1,B_2 \in \mathcal{B}\), we also have \(B_1,B_2 \in \tau\). Since \(\tau\) is closed under intersections, we have \(B_1 \cap B_2 \in \tau\). Since \(\mathcal{B}\) generates \(\tau\), we have \(B_1 \cap B_2\) is a union of members of \(\mathcal{B}\).

To prove the other way around, that is, if \(X = \bigcup_{B \in \mathcal{B}} B\) and for any \(B_1,B_2 \in \mathcal{B}\), the set \(B_1 \cap B_2\) is a union of members of \(\mathcal{B}\), then \(\mathcal{B}\) is a basis for the topology on \(X\). Let \(\tau(\mathcal{B}) = \{\bigcup_{\alpha} B_{\alpha}: B_{\alpha} \in \mathcal{B}\}\). We need to prove that \(\tau\) is a topology.

1. Clearly, we have \(X, \emptyset \in \tau\). (This follows from the condition in the problem and from the fact that the empty set is a union of no sets.)
2. Any arbitrary union of elements in \(\tau\) is nothing but an arbitrary union of elements in \(\mathcal{B}\) which is again in \(\tau\).
3. Consider two elements in \(\tau\). These two elements can be written as \(\bigcup_{\alpha} B^1_{\alpha}\) and \(\bigcup_{\beta} B^2_{\beta}\) where \(B^1_{\alpha},B^2_{\beta} \in \mathcal{B}\) forall \(\alpha,\beta\). The intersection of these two elements is given by \(\left(\bigcup_{\alpha} B^1_{\alpha} \right) \cap \left(\bigcup_{\beta} B^2_{\beta} \right) = \bigcup_{\alpha,\beta} \left(B^1_{\alpha} \cap B^2_{\beta} \right)\). But we are given that intersection of two elements in \(\mathcal{B}\) can be written as union of members of \(\mathcal{B}\) i.e. \(B^1_{\alpha} \cap B^2_{\beta} = \bigcup_{\gamma} B_{\gamma,\alpha,\beta}\). Hence, the intersection of any two elements in \(\tau\) is given by \(\bigcup_{\alpha,\beta,\gamma} B_{\gamma,\alpha,\beta}\). Hence, closed under finite intersections.

Hence, \(\tau(\mathcal{B})\) is a topology.

Elementary Number Theory - 1.1 Definitions and Propositions

Definition:

Suppose that \(a,b \in \mathbb{Z}\) and \(a\neq 0\). Then we say that \(a\) divides \(b\), denoted by \(a|b\), if there exists \(c \in \mathbb{Z}\) such that \(b=ac\). In this case, we also say that \(a\) is a divisor of \(b\) (or) \(b\) is a multiple of \(a\).

Theorem:

Suppose that \(a \in \mathbb{N}\) and \(b \in \mathbb{Z}\). Then there exists unique \(q,r \in \mathbb{Z}\) such that \(b=aq+r\) and \(0 \leq r \lt a\)

Proof:

First we need to show the existence of \(q,r \in \mathbb{Z}\). Consider the set \[S = \{b-as \geq 0 : s \in \mathbb{Z}\}.\] \(S \subset \mathbb{N}\) and hence by well-ordering principle there exists a least element in \(S\). Call the least element \(r\). Claim now is that \(0 \leq r \lt a\). Clearly, since \(r \in S\), we have \(r \geq 0\). Now assume that \(r \geq a\). Now \(r = b-as_1 \geq a\) for some \(s_1 \in \mathbb{Z}\). Consider \(r_1 = r-a = b- as_1 - a \geq a-a = 0\) i.e. \(r_1 = b- a (s_1 + 1) \geq 0 \implies r_1 \in S\). However, \(r_1 < r\). This contradicts the minimality of \(r\). Hence, \(0 \leq r \lt a\) and thereby the existence follows.

To prove the uniqueness, as always, assume that there are two distinct sets of numbers \(q_1,r_1 \in \mathbb{Z}\) and \(q_2,r_2 \in \mathbb{Z}\) such that \(aq_1 + r_1 = b = aq_2 + r_2\) and \(0 \leq r_1,r_2 \lt a\). Assume without loss of generality that \(q_1 \lt q_2\). This obviously implies \(r_1 \gt r_2\). Hence, we get \(0 \lt a(q_2-q_1) = r_1 - r_2\). This means that \(a|(r_1 - r_2)\). But, since \(r_1 \gt r_2\), \(0 \lt r_1 - r_2 \lt a\). CONTRADICTION. Hence, the uniqueness follows. \(\Box\)

Theorem:

Suppose that \(a,b \in \mathbb{N}\). Then there exists a unique \(d \in \mathbb{N}\) such that

1. there exists \(x,y \in \mathbb{Z}\) such that \(d = ax+by\);
2. \(d|a\) and \(d|b\);
3. for every \(k\in \mathbb{N}\) such that \(k|a\) and \(k|b\), we have \(k|d\).

Proof:

Consider the set \(I = \{au + bv : u,v \in \mathbb{Z}\}\). Clearly, \(I \subseteq \mathbb{Z}\). By well-ordering principle, there exists a least positive element in \(I\). Call it \(d\). We have \(d = ax+by\) for some \(x,y \in \mathbb{Z}\). 

We now want to show that \(d|a\) and \(d|b\). In fact, we shall show that \(d|(au+bv)\), \(\forall u,v, \in \mathbb{Z}\). By the previous theorem, we get that \(au+bv = dq + r\) where \(0 \leq r \lt d\). Since \(d = ax+by\), we get \(au+bv - (ax+by)q = r\) i.e. \(r = a(u-qx) + b(v-qy)\). Hence, \(r \in I\) and \(0 \leq r \lt d\). But \(d\) is the least positive element in \(I\). This gives us that \(r=0\). Hence, \(d\) divides all elements in the set \(I\). Setting \(u=1\) and \(v=0\) gives us that \(d|a\) and setting \(u=0\) and \(v=1\) gives us that \(d|b\).

We now need to prove that for every  \(k\in \mathbb{N}\) such that \(k|a\) and \(k|b\), we have \(k|d\). We have that \(d = ax+by\). Further since \(k|a\) and \(k|b\), we get \(a=ka_1\) and \(b=kb_1\), where \(a_1,b_1 \in \mathbb{Z}\). Hence, we get \(d = k (a_1x + b_1y)\). Since \(a_1x + b_1y \in \mathbb{Z}\), we get \(k|d\).  \(\Box\)

Definition:

The number \(d\) satisfying the criteria in the above theorem is called the greatest common divisor of \(a\) and \(b\). It is typically denoted as \(d=(a,b)\)

Theorem: (Euclid Algorithm)

Suppose that \(a,b \in \mathbb{N}\). Suppose further that \(q_1,q_2,\ldots,q_{n+1} \in \mathbb{Z}\) and \(r_1,r_2,\ldots,r_n \in \mathbb{N}\) satisfy \(0 \lt r_n \lt r_{n-1} \lt \cdots \lt r_1 \lt a\) and \[b = aq_1 + r_1\] \[a = r_1q_2 + r_2\] \[r_1 = r_2q_3 + r_3\] \[\vdots\] \[r_{n-2} = r_{n-1}q_n + r_n\] \[r_{n-1} = r_{n}q_{n+1}\] Then \((a,b) = r_n\)

Proof:

We shall first prove that \((b,a) = (a,r_1)\). This is not hard to prove. This follows immediately once we note that if \(k\) divides \(a\) and \(b\) then \(k\) has to divide \(r_1\). Similarly, if \(m\) divides \(a\) and \(r_1\) then \(m\) has to divide \(b\). Hence, \((b,a) = (a,r_1)\). Now it follows by repeatedly recognizing this fact that \((b,a) = (a,r_1) = (r_1,r_2) = (r_2,r_3) = \cdots = (r_{n-2},r_{n-1}) = (r_{n-1},r_{n}) = (q_{n+1}r_{n},r_{n}) = r_n\). \(\Box\)

Theorem:

Suppose that \(a,b \in \mathbb{N}\) and \((a,b) = 1\). Suppose further that \(w \in \mathbb{N}\) satisfies \(w|ab\), then there exists a unique \(u,v \in \mathbb{N}\) such that \(u|a\), \(v|b\) and \(w = uv\)

Proof:

Let \(u = (w,a)\) and \(v = (w,b)\). We shall show that \(u|a\), \(v |b\) and \(w = uv\). From the definition of \(u\) and \(v\) it is clear that \(u|a\) and \(v|b\). All we need to show is that \(uv = w\). From the theorem proved earlier, we have that \(u = wx_1 + ay_1\) for some \(x_1,y_1 \in \mathbb{Z}\) and \(v = wx_2 + by_2\) for some \(x_2,y_2 \in \mathbb{Z}\). Hence, we get \(uv = (wx_1 + ay_1)(wx_2 + by_2)\). We also have \(ab = wk\) for some \(k \in \mathbb{Z}\). Hence expanding we get, \(uv = w(wx_1x_2 + x_1by_2 + ay_1x_2 + ky_1y_2)\). Hence, \(w|(uv)\).

Now we will show that \((uv)|w\). Since \((a,b) = 1\), \(\exists x,y \in \mathbb{Z}\) such that \(ax+by = 1\). Multiply by \(w\) to get \(w = wax + wby\). Since \(u = (w,a)\), we get \(w = uw_u\) and \(a = ua_u\). Also since \(v = (w,b)\), we get \(w = v w_v\) and \(b = v b_v\). Hence, \(w = v w_v u a_u x + u w_u v b_v y \implies w = u v (w_v a_u x + w_u b_v y) \implies (uv)|w\).

Now, we still need to show uniqueness. If we have another set, say \(u_1,v_1\), such that \(u_1 | a\), \(v_1 | b\) and \(w = u_1 v_1\). We have that \(u = wx_1 + ay_1 = u_1 v_1 x_1 + u_1 k_1 y_1 = u_1 (v_1 x_1 + k_1 y_1)\) . Hence, we get \(u_1|u\) i.e. \(u = u_1 m_1\). Similarly, we have that \(v = wx_2 + by_2 = u_1 v_1 x_2 + v_1 k_2 y_2 = v_1 (u_1 x_2 + k_2 y_2)\) . Hence, we get \(v_1|v\) i.e. \(v = v_1 l_1\). But we have that \(uv = w = u_1v_1\) and hence we get \(l_1m_1 = 1\) and since \(u,v,u_1,v_1 \in \mathbb{N}\), we get \(l_1 = m_1 = 1\) and hence \(u_1 = u\) and \(v_1 = v\). \(\Box\)

Topology without Tears - 2.1 Definitions and Propositions

Definition:

A subset \(S\) of \(\mathbb{R}\) is said to be open in the euclidean topology on \(\mathbb{R}\) if it has the following property:

For each \(x \in S\), there exists \(a,b \in \mathbb{R}\), with \(a \lt b\) such that \(x \in (a,b) \subseteq S\).

Notation: Whenever we refer to the topological space \(\mathbb{R}\) without specifying the topology, we mean \(\mathbb{R}\) with the euclidean topology.

Proposition:

The euclidean topology is indeed a topology on \(\mathbb{R}\)

Proof:

The drill again.

• Clearly, \(\mathbb{R} \in \tau\) since given any \(x \in \mathbb{R}\), the interval \((x-1,x+1) \subset \mathbb{R}\). \(\emptyset \in \tau\) since the statement "for each \(x \in \emptyset\), there exists \(a,b \in \mathbb{R}\), with \(a \lt b\) such that \(x \in (a,b) \subseteq \emptyset\)" is vacuously true.
• Let \(A_{\alpha} \in \tau\) for every \(\alpha \in \Gamma\). Let \(A = \bigcup_{\alpha \in \Gamma} A_{\alpha}\). Consider \(x \in A\). This means \(x \in A_{\alpha}\) for some \(\alpha \in \Gamma\). Since \(A_{\alpha} \in \tau\), we can find an interval say \((a,b)\) containing \(x\) such that \((a,b) \subseteq A_{\alpha}\). But we have \(A_{\alpha} \subseteq A\) and hence \((a,b) \subseteq A\). Hence, for each \(x \in A\), there exists \(a,b \in \mathbb{R}\) with \(a \lt b\) such that \(x \in (a,b) \subseteq A\). Hence, \(\tau\) is closed under arbitrary unions.
• Let \(A_1,A_2 \in \tau\). Let \(A = A_1 \cap A_2\). Consider \(x \in A\). This means we also have \(x \in A_1\) and \(x \in A_2\) where \(A_1,A_2 \in \tau\). Hence, there exists an interval \((a_1,b_1)\) containing \(x\) such that \((a_1,b_1) \subseteq A_1\) and an interval \((a_2,b_2)\) containing \(x\) such that \((a_2,b_2) \subseteq A_2\). Now let \(a = \max(a_1,a_2)\) and \(b = \min(b_1,b_2)\). Note that \(x \in (a,b)\). We also have \((a,b) \subseteq (a_1,b_1) \subseteq A_1\) and \((a,b) \subseteq (a_2,b_2) \subseteq A_2\). Hence, there exists \(a,b \in \mathbb{R}\), with \(a \lt b\), such that \(x \in (a,b) \subseteq A_1 \cap A_2\). Hence \(\tau\) is closed under finite intersections. \(\Box\)

Proposition:

Let \(r,s \in \mathbb{R}\) with \(r \lt s\). In the euclidean topology \(\tau\) on \(\mathbb{R}\), the open interval \((r,s)\) does indeed belong to \(\tau\) and so is an open set.

Proof:

Given any \(x \in (r,s)\), we want to find an open interval containing \(x\) and lying within \(r,s\). However note that we have \(x \in (r,s) \subseteq (r,s)\). Hence, \((r,s)\) is indeed an open set. \(\Box\)

Proposition:

The open intervals \((r, \infty)\) and \((-infty,r)\) are both open sets in \(\mathbb{R}\), for every real number \(r\).

Proof:

1. The open interval \((r, \infty)\) is an open set. Consider any \(x \in (r,\infty)\). Consider the open interval \((r,x+1)\). Clearly, we have \(x \in (r,x+1) \subset (r,\infty)\). Hence, the open interval \((r, \infty)\) is an open set.
2. The open interval \((-\infty,r)\) is an open set. Consider any \(x \in (-\infty,r)\). Consider the open interval \((x-1,r)\). Clearly, we have \(x \in (x-1,r) \subset (-\infty,r)\). Hence, the open interval \((-\infty,r)\) is an open set. \(\Box\)

Proposition:

For each \(c,d \in \mathbb{R}\) with \(c \lt d\), the closed interval \([c,d]\) is not an open set in \(\mathbb{R}\). In fact, the closed interval \([c,d]\) is a closed set in \(\mathbb{R}\).

Proof:

Consider the complement of \([c,d]\). We have \(\mathbb{R} \backslash [c,d] = (-\infty,c) \cup (d, \infty)\). From previous part, we have that \((-\infty,c)\) and \((d,\infty)\) are open sets. And union of open sets is also open and hence \(\mathbb{R} \backslash [c,d] \) is open. Hence, \([c,d]\) is a closed set. We still need to prove that it is not open. Consider the point \(c \in [c,d]\). Suppose that there exists \(a,b \in \mathbb{R}\) such that \(c \in (a,b)\). We will prove that \((a,b) \not\subseteq [c,d]\). To see this, consider the point \(\frac{a+c}{2}\). We have \(a \lt \frac{a+c}{2} \lt c\). Hence, we have that \(\frac{a+c}{2} \in (a,b)\) but \(\frac{a+c}{2} \notin [c,d]\). Hence, \((a,b) \not\subseteq [c,d]\). CONTRADICTION. Hence, \([c,d]\) is not an open set. \(\Box\)

Proposition:

Every singleton set \(\{a\}\) is closed in \(\mathbb{R}\).

Proof:

One way to think of \(\{a\}\) is nothing but the closed interval \([a,a]\). The proof is same as the one for the previous case. There is no difference in the proof if \(a \lt b\) (or) \(a=b\).

Proposition:

The set \(\mathbb{Z}\) of all integers is a closed subset of \(\mathbb{R}\).

Proof:

Consider the complement of \(\mathbb{Z}\) in \(\mathbb{R}\) i.e. \(A = \mathbb{R} \backslash \mathbb{Z}\). Now consider any point \(x \in A\). Let \(\epsilon = \min(\{x-\lfloor x \rfloor , \lceil x \rceil - x\})\). Note that \(\epsilon > 0\) since \(x\) is never an integer. Consider the open interval \((x-\epsilon,x+\epsilon)\). Hence, for every \(x \in \mathbb{R} \backslash \mathbb{Z}\), we can find \((x-\epsilon,x+\epsilon) \in \mathbb{R} \backslash \mathbb{Z}\). Hence, the set \(\mathbb{R} \backslash \mathbb{Z}\) is open. This gives us that the set \(\mathbb{Z}\) is a closed set.

Proposition:

The set \(\mathbb{Q}\) of all rational numbers is neither a closed subset of \(\mathbb{R}\) nor an open subset of \(\mathbb{R}\).

Proof:

Let \(\mathbb{I} = \mathbb{R} \backslash \mathbb{Q}\) be the set of irrationals.

1. We shall first prove that \(\mathbb{Q}\) is not open under the euclidean topology. To see this, we shall prove it by contradiction. Assume that the set is open i.e. given any point \(x \in \mathbb{Q}\), there exists an open interval of the form \((a,b) \subseteq \mathbb{Q}\). Let \(\epsilon = \min(\{x-a,b-x\})\). By Archimedean property, there exists an \(n \in \mathbb{N}\) such that \(\frac{\sqrt{2}}{n} < \epsilon\). This gives us that \(x \pm \frac{\sqrt{2}}{n} \in (a,b)\). However note that \(x \pm \frac{\sqrt{2}}{n} \notin \mathbb{Q}\). This gives us a CONTRADICTION. Hence, \(\mathbb{Q}\) is not open under the euclidean topology.
2. We shall now prove that \(\mathbb{Q}\) is not closed. To prove this, we shall prove that \(\mathbb{I}\) is not open. We shall prove this by contradiction. Assume that \(\mathbb{I}\) is open i.e. given any point \(x \in \mathbb{I}\), there exists an open interval of the form \((a,b) \subseteq \mathbb{I}\). Let \(\epsilon = \min(\{x-a,b-x\})\). There exists \(n \in \mathbb{N}\) such that \(\frac1{10^n} < \epsilon\). Assuming \(x\) is expressed in decimal system, for any \(x\), we get \(\displaystyle x - \frac{\lfloor 10^{n} x \rfloor}{10^{n}} \lt \frac1{10^n} \lt \epsilon\). But note that \(y = \frac{\lfloor 10^{n} x \rfloor}{10^{n}}\) is a rational number and \(y \in (a,b)\). CONTRADICTION. Hence, \(\mathbb{I}\) is not open under the euclidean topology. Hence, \(\mathbb{Q}\) is not closed under the euclidean topology.

Hence, \(\mathbb{Q}\) is neither open nor closed.

Topology without Tears - 2.1 Exercises

Exercise:

1. Prove that if \(a,b \in \mathbb{R}\) with \(a \lt b\), then neither \([a,b)\) nor \((a,b]\) is an open subset of \(\mathbb{R}\). Also show that neither is a closed subset of \(\mathbb{R}\).
• We shall first consider the interval \([a,b)\).
• We will prove that the interval \([a,b)\) is not open. If \([a,b)\) is open, then \(\forall x \in [a,b)\), we have an interval \((c,d)\) containing \(x\) such that \((c,d) \subseteq [a,b)\). We will show that the point \(a\) doesn't lie in any open interval \((c,d)\) containing \(a\) such that \((c,d) \subseteq [a,b)\). The proof goes by contradiction. If there exists \((c,d)\) containing \(a\), such that \((c,d) \subseteq [a,b)\), consider \(x = \frac{a+c}{2}\). Note that \(x \in (c,d)\) but \(x \notin [a,b)\). Hence, \((c,d) \not\subseteq [a,b)\) for any open interval \((c,d)\). Hence, \([a,b)\) is not open.
• We will now prove that the interval \([a,b)\) is not closed. Look at the complement in \(\mathbb{R}\) and the argument is similar as above to prove that the complement in \(\mathbb{R}\) is not open as well. Hence, \([a,b)\) is not closed.
• Now consider the interval \((a,b]\) and the argument is almost the same as above.
2. Prove that the sets \([a,\infty)\) and \((-\infty,a]\) are closed subsets of \(\mathbb{R}\).
• Consider \(\mathbb{R} \backslash [a,\infty) = (-\infty,a)\). Consider any \(x \in (-\infty,a)\). Let \(\epsilon = a-x\) and consider the subset \((x-\epsilon,x+\epsilon)\). Clearly, \(x \in (x-\epsilon,x+\epsilon) \subset (-\infty,a)\). This is true for any \(x \in (-\infty,a)\). Hence, \((-\infty,a)\) is an open subset. Hence, \([a,\infty)\) is a closed set of \(\mathbb{R}\).
• Consider \(\mathbb{R} \backslash (-\infty,a] = (a,\infty)\). Consider any \(x \in (a,\infty)\). Let \(\epsilon = x-a\) and consider the subset \((x-\epsilon,x+\epsilon)\). Clearly, \(x \in (x-\epsilon,x+\epsilon) \subset (a,\infty)\). This is true for any \(x \in (a,\infty)\). Hence, \((a,\infty)\) is an open subset. Hence, \((-\infty,a]\) is a closed set of \(\mathbb{R}\).
3. Show, by example, that the union of an infinite number of closed subsets of \(\mathbb{R}\) is not necessarily a closed subset of \(\mathbb{R}\).
• Let \(A_n = [\frac1n,1-\frac1n]\). Then, \(A = \bigcup_{n=2}^{\infty} A_n = \bigcup_{n=2}^{\infty} [\frac1n,1-\frac1n] = (0,1)\) which is clearly an open set of \(\mathbb{R}\).
4. Prove each of the following statements.
• The set \(\mathbb{Z}\) of all integers is not an open subset of \(\mathbb{R}\).
• Proof by contradiction. Assume that it is an open subset, then for each \(x \in \mathbb{Z} \), there exists an open interval, say \((a,b)\) such that \(x \in (a,b)\) and \((a,b) \subseteq \mathbb{Z}\). Let \(\epsilon = \min(x-a,b-x)\). By Archimedean property, choose a positive integer \(n \geq 2\) such that \(\frac1n \lt \epsilon\). Then we have that \(x \pm \frac1n \in (a,b)\) but \(x \pm \frac1n\) is not an integer. Hence, \(\mathbb{Z}\) of all integers is not an open subset of \(\mathbb{R}\).
• The set \(\mathcal{P}\) of all prime numbers is a closed subset of \(\mathbb{R}\) but not an open subset of \(\mathbb{R}\).
• We shall first prove that \(\mathcal{P}\) is a closed subset of \(\mathbb{R}\). Consider \(\mathcal{C} = \mathbb{R} \backslash \mathcal{P}\). We will prove that this set is open. Note that since primes are a subset of natural numbers, it is well-ordered. In particular, we can write \(\mathcal{P} = \{p_k \in \mathbb{N} : p_k \{ is the \}k^{th} \text{ prime}\}\) i.e. we have \(p_1 \lt p_2 \lt p_3 \lt \cdots\). This means \(\mathcal{C} = (-\infty,p_1) \bigcup_{k=1}^{\infty} (p_k,p_{k+1})\). Note that \((-\infty,p_1)\) and \((p_k,p_{k+1})\) are open sets. Since union of open sets is still an open set, we get that \(\mathcal{C}\) is an open set. Hence, we have that \(\mathcal{P}\) is a closed subset of \(\mathbb{R}\).
• We shall now prove that \(\mathcal{P}\) is not an open subset of \(\mathbb{R}\). The proof goes by contradiction. Assume that it is an open subset, then given any \(p_k \in \mathcal{P}\), there exists and open subset \((a,b) \subset \mathcal{P}\) such that \(p_k \in (a,b)\). Let \(\epsilon = \min(\{p_k-a,b-p_k\})\). By Archimedean property, we have that there exists a positive integer \(n \geq 2\) such that \(\frac1n \lt \epsilon\). Now we have \(p_k \pm \frac1n \in (a,b)\) but \(p_k \pm \frac1n \notin \mathcal{P}\). CONTRADICTION. Hence, \(\mathcal{P}\) is not an open subset of \(\mathbb{R}\).
• The set \(\mathbb{I}\) of all irrational numbers is neither a closed subset nor an open subset of \(\mathbb{R}\).
• Note that if \(A\) is not open, then \(X \backslash A\) is not closed and similarly if \(A\) is not closed, then \(X \backslash A\) is not open. We proved as a proposition earlier that the set \(\mathbb{Q} = \mathbb{R} \backslash \mathbb{I}\) is neither open nor closed. Hence, \(\mathbb{I}\) is neither open nor closed.
5. If \(F\) is a non-empty finite subset of \(\mathbb{R}\), then show that \(F\) is closed in \(\mathbb{R} \) but that \(F\) is not open in \(\mathbb{R}\).
• Since \(F\) is a non-empty finite subset of \(\mathbb{R}\), we can write \(F = \{f_1,f_2,\ldots,f_n\}\) where \(n \in \mathbb{Z}^+\) and \(f_1 \lt f_2 \lt \cdots \lt f_n\).
• We shall first prove that \(F\) is closed. Let \(F^c = \mathbb{R} \backslash F\). Then we have \(F^c = (-\infty,f_1) \cup \left( \bigcup_{k=1}^{n-1} (f_k,f_{k+1})\right)\). Each of the set i.e. \((-\infty,f_1)\) and \((f_k,f_{k+1})\) are open sets. Hence, union of open sets is again open. Hence, \(F^c\) is open. Hence, \(F\) is a closed set of \(\mathbb{R}\).
• We shall not prove that \(F\) is not open in \(\mathbb{R}\). We shall prove by contradiction. Assume \(F\) is open. This means that for any \(f_k \in F\), we have an open set say \((a,b)\) such that \(f_k \in (a,b)\) and \((a,b) \subseteq F\). Let \(\epsilon = \min(\{f_k-a,b-f_k,f_{k+1}-f_k,f_k - f_{k-1}\})\). By Archimedean property, we can choose a positive integer \(n \geq 2\), such that \(\frac1n \lt \epsilon\). Consider \(f_k \pm \frac1n\). We have \(f_k \pm \frac1n \in (a,b)\) but \(f_k \pm \frac1n \notin F\). CONTRADICTION. Hence, \(F\) is not an open subset of \(\mathbb{R}\).
6. If \(F\) is a non-empty countable subset of \(\mathbb{R}\), prove that \(F\) is not an open set.
• \(F\) is a countable subset of \(\mathbb{R}\). We shall prove by contradiction that \(F\) is not open. Assume the \(F\) is a open subset of \(\mathbb{R}\). Then this means that for every \(x \in F\), there exists an open interval \((a,b) \subseteq F\). Hence, we now have \((a,b) \subseteq F\). However note that \((a,b)\) is an uncountable set whereas \(F\) is a countable set. But any subset of a countable set is again a countable set. CONTRADICTION. Hence, \(F\) is not an open set.
7. 
   
• Let \(S = \{0,1,\frac12,\frac13,\ldots,\frac1n,\ldots\}\). Prove that the set \(S\) is closed in the euclidean topology on \(\mathbb{R}\).
• Let \(S^c = \mathbb{R} \backslash S\). Hence, we have \(S^c = \left( \bigcup_{k=1}^{\infty} \left(\frac1{k+1},\frac1k \right) \right) \cup (-\infty,0) \cup (1,\infty)\). Note that each of the interval in the union is an open interval. Hence, the set \(S^c\) is also an open set. Hence, \(S\) is a closed set on \(\mathbb{R}\).
• Is the set \(T = \{1,\frac12,\frac13,\ldots,\frac1n,\ldots\}\) closed in \(\mathbb{R}\)?
• Let \(T^c = \mathbb{R} \backslash T\). Hence, we have \(T^c = \left( \bigcup_{k=1}^{\infty} \left(\frac1{k+1},\frac1k \right) \right) \cup (-\infty,0] \cup (1,\infty)\). We shall prove that \(T^c\) is not open. Again, the proof is by contradiction. Assume that the set \(T^c\) is open. Consider the point \(0\). Since \(T^c\) is assumed to be open, there exists an open set \((a,b)\) containing \(0\) and \((a,b) \subseteq T^c\). Let \(\epsilon = \min(-a,b)\). By Archimedean property, there exists an integer \(n \geq 2\) such that \(\frac1n \lt \epsilon\). Hence, \(\frac1n \in (a,b)\). However, \(\frac1n \notin T^c\). This CONTRADICTS the fact the \((a,b) \subseteq T^c\). Hence, \(T^c\) is not open. Hence, \(T\) is not closed.
• Is the set \(S = \{\sqrt{2},2 \sqrt{2}, 3 \sqrt{2}, \ldots, n \sqrt{2}, \ldots\}\) closed in \(\mathbb{R}\)?
• We have \(S^c = \left( \bigcup_{k=1}^{\infty} (k\sqrt{2}, (k+1) \sqrt{2}) \right) \cup (-\infty,\sqrt{2})\). \(S^c\) is a union of open intervals which are open sets. Hence, \(S^c\) is a open set. Hence, \(S\) is a closed set in \(\mathbb{R}\).
8. 
   
• Let \((X,\tau)\) be a topological space. A subset \(S\) of \(X\) is said to be an \(F_{\sigma}\)-set if it is the union of a countable number of closed sets. Prove that all open intervals \((a,b)\) and all closed intervals \([a,b]\), are \(F_{\sigma}\)-sets in \(\mathbb{R}\).
• We have \((a,b) = \bigcup_{k=1}^{\infty} \left[a+\frac1k,b- \frac1k \right] \). Hence, any open interval is a \(F_{\sigma}\) set.
• We have \([a,b] = \bigcup_{k=1}^{\infty} [a,b] \). Hence, any closed interval is a \(F_{\sigma}\) set.
• Let \((X,\tau)\) be a topological space. A subset \(S\) of \(X\) is said to be an \(G_{\delta}\)-set if it is the intersection of a countable number of open sets. Prove that all open intervals \((a,b)\) and all closed intervals \([a,b]\), are \(G_{\delta}\)-sets in \(\mathbb{R}\).
• We have \((a,b) = \bigcap_{k=1}^{\infty} (a,b)\). Hence, any open interval is a \(G_{\delta}\) set.
• We have \([a,b] = \bigcap_{k=1}^{\infty} \left(a -\frac1k, b + \frac1k \right) \). Hence, any closed interval is a \(G_{\delta}\) set.
• Prove that the set of rationals is an \(F_{\delta}\) set in \(\mathbb{R}\).
• The rationals form a countable set. Hence, we can list the rationals as \(\mathbb{Q} = \{q_k\}_{k=1}^{\infty}\). We proved in one of the propositions earlier that the set \(\{a\}\) is a closed set in \(\mathbb{R}\). Hence, we have \(\mathbb{Q} = \bigcup_{k=1}^{\infty} \{q_k\}\) which means \(\mathbb{Q}\) is a countable union of closed sets in \(\mathbb{R}\). Hence, the set of rationals is an \(F_{\delta}\) set in \(\mathbb{R}\).
• Verify that the complement of an \(F_{\sigma}\)-set is a \(G_{\delta}\)-set and the complement of a \(G_{\delta}\)-set is a \(F_{\sigma}\)-set.
• First we shall prove that the complement of an \(F_{\sigma}\)-set is a \(G_{\delta}\)-set. Let \(A\) be an \(F_{\sigma}\)-set. This means that \(A = \bigcup_{k=1}^{\infty} F_{k}\) where \(F_k\) are closed sets. This means that \(F_k^c\) are open sets. We have \(A^c = \bigcap_{k=1}^{\infty} F_k^c\), where \(F_k^c\) are open sets and hence \(A^c\) is a countable intersection of open sets. Hence, \(A^c\) is a \(G_{\delta}\)-set.
• Now we shall prove that the complement of a \(G_{\delta}\)-set is a \(F_{\sigma}\)-set. Let \(A\) be a \(G_{\delta}\)-set. This means that \(A = \bigcap_{k=1}^{\infty} G_k\) where \(G_k\) are open sets. This means that \(G_k^c\) are closed sets. We have \(A^c = \bigcup_{k=1}^{\infty} G_k^c\), where \(G_k^c\) are closed sets and hence \(A^c\) is a countable union of closed sets. Hence, \(A^c\) is a \(F_{\sigma}\)-set.

Topology without Tears - 1.3 Exercises

Exercises:

1. Let \(f\) be a function from a set \(X\) into a set \(Y\). Then \[f^{-1}\left( \bigcup_{j \in J} B_j\right) = \bigcup_{j \in J} f^{-1} \left(B_j\right)\] and \[f^{-1}\left( B_1 \bigcap B_2 \right) = f^{-1}\left(B_2\right) \bigcap f^{-1}\left(B_2\right)\] for any subsets \(B_j\) of \(Y\) and any index set \(J\).
• Prove the above two expressions are true
• These have already been proved as claims as part of the previous propositions
• Find sets \(A_1,A_2,X\) and \(Y\) and a function \(f: X \rightarrow Y\) such that \(f \left(A_1 \bigcap A_2 \right) \neq f(A_1) \bigcap f(A_2)\) where \(A_1,A_2 \subset X\)
• Let \(X = \{1,2,3\}\). Let \(A_1 = \{1,2\}\) and \(A_2 = \{2,3\}\). Define the function \(f\) on \(X\) as follows \[f(1) = 5, f(2) = 4, f(3) = 5\]. We have \(f(A_1) = \{4,5\}\), \(f(A_2) = \{4,5\}\) and hence \(f(A_1) \bigcap f(A_1) = \{4,5\}\). Further, \(A_1 \cap A_2 = \{2\}\). Hence, \(f(A_1 \cap A_2) = \{4\}\). Clearly, we see that \[f \left( A_1 \bigcap A_2\right) = \{4\} \neq \{4,5\} = f(A_1) \bigcap f(A_2)\]
2. Is the topology described in problem \(6 (ii)\) of the first exercises a finite closed topology?
• \(X = \mathbb{N}\) and \(\tau = \{\{n,n+1,\ldots\}: n \in \mathbb{N}\}\). From the definition of finite closed topology, we need all the finite subsets of \(X\) to be closed. Consider \(A = \{3\}\). This set is clearly not open. Further, \(A^c = \{0,1,2,4,5,6,\ldots\} \notin \tau\). Hence, this is set is not closed as well. So the topology is not a finite-closed topology.
3. A topological space \(X,\tau\) is said to be a \(T_1\)-space if every singleton set \{x\} is closed in \(X,\tau\). Show that precisely two of the following nine topological spaces are \(T_1\)-spaces. (Justify your answer.)
• A discrete space: In a discrete space every subset is clopen. Hence, every singleton set is closed and hence the space is \(T_1\).
• An indiscrete space with at-least two points: The indiscrete space has only \(\emptyset,X\) as clopen sets and the remaining subsets of \(X\) are neither open nor closed. Hence the space is not \(T_1\).
• An infinite set with the finite-closed topology: A co-finite topology has only \(X\) and all finite sets as its closed sets. Since every singleton set is definitely a finite set, the space is \(T_1\).
• \(X = \{a,b,c,d,e,f\}\) and \(\tau = \{X, \emptyset,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e,f\}\}\): The set \(\{b\}\) for instance is neither open nor closed. Hence, the space is not \(T_1\).
• \(X = \mathbb{R}\) and \(\tau\) consists of \(\mathbb{R}, \emptyset\) and every interval \((-n,n)\), for any positive integer: The set \(\{2\}\) for instance is neither open nor closed. Hence, the space is not \(T_1\).
• \(X = \mathbb{R}\) and \(\tau\) consists of \(\mathbb{R}, \emptyset\) and every interval \([-n,n]\), for any positive integer: The set \(\{2\}\) for instance is neither open nor closed. Hence, the space is not \(T_1\).
• \(X = \mathbb{R}\) and \(\tau\) consists of \(\mathbb{R}, \emptyset\) and every interval \([n,\infty)\), for any positive integer: The set \(\{2\}\) for instance is neither open nor closed. Hence, the space is not \(T_1\).
• \(X = \mathbb{N}\) and \(\tau\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{1,2,\ldots,n\}\) for any positive integer \(n\): The set \(\{2\}\) for instance is neither open nor closed. Hence, the space is not \(T_1\).
• \(X = \mathbb{N}\) and \(\tau\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{n,n+1,\ldots\}\) for any positive integer \(n\): The set \(\{2\}\) for instance is neither open nor closed. Hence, the space is not \(T_1\).
4. Let \(\tau\) be the finite-closed topology on a set \(X\). It \(\tau\) is also the discrete topology, prove that the set \(X\) is finite.
• Since \(\tau\) is a co-finite topology, we have that all the sets in \(\tau\) to be co-finite except probably the empty-set. We also have that \(\tau\) is a discrete topology which means \(\tau\) contains all the subsets of \(X\). In particular, \(\tau\) contains singleton sets i.e. say if \(x \in X\) then \(\tau\) contains \(\{x\}\). \(\tau\) is co-finite implies \(X \backslash x\) is finite. Hence, \(X = \{x\} \bigcup \left(X \backslash x \right)\) is again finite.
5. A topological space \((X, \tau)\) is said to be a \(T_0\)-space if for each pair of distinct points \(a,b \in X\), either there exists an open set containing \(a\) and not \(b\), or there exists an open set containing \(b\) and not \(a\).
• Prove that every \(T_1\)-space is a \(T_0\)-space
• A \(T_1\) space is one in which every singleton set is a closed set. Consider two points \(a,b \in X\). The sets \(X \backslash \{a\}\) and \(X \backslash \{b\}\) are open since the space is \(T_1\). Further, \(b \in X \backslash \{a\}\) and also \(a \in X \backslash \{b\}\). Hence, in-fact in a \(T_1\) space we have that for each pair of distinct points \(a,b \in X\), either there exists an open set containing \(a\) and not \(b\), and there exists an open set containing \(b\) and not \(a\). Hence, \(T_1\)-space is a \(T_0\)-space.
• A discrete space is a \(T_0\) space since it is a \(T_1\) space.
• An indiscrete space with at least two points is not a \(T_0\) space since if we take any element in the set \(X\), there is no open set containing only one of them and not the other one.
• An infinite set with a co-finite topology is a \(T_0\) space since it is a \(T_1\) space.
• \(X = \{a,b,c,d,e,f\}\) and \(\tau = \{X, \emptyset,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e,f\}\}\) is not \(T_0\) since if consider \(c,d \in X\), there is no open set containing only one of them and not the other one.
• \(X = \mathbb{R}\) and \(\tau\) consists of \(\mathbb{R}, \emptyset\) and every interval \((-n,n)\), for any positive integer.
• Consider the elements \(1.5,1.6 \in X\). There is no open set containing only one and not the other. Hence the space is not \(T_0\).
• \(X = \mathbb{R}\) and \(\tau\) consists of \(\mathbb{R}, \emptyset\) and every interval \([-n,n]\), for any positive integer.
• Consider the elements \(1.5,1.6 \in X\). There is no open set containing only one and not the other. Hence the space is not \(T_0\).
• Find a topology \(\tau\) on the set \(X = \{0,1\}\) so that \((X,\tau)\) will be a \(T_0\)-space but not a \(T_1\)-space. (This topological space is called the Sierpinski space.)
• Let \(\tau = \{\emptyset,\{0\},X\}\). This space is a \(T_0\)-space but not a \(T_1\)-space.
• \(X = \mathbb{N}\) and \(\tau\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{1,2,\ldots,n\}\) for any positive integer \(n\). This space is a \(T_0\) space since given any two distinct elements say \(m,n \in \mathbb{N}\), we can consider the set \(\{1,2,\ldots,\min(m,n)\}\). Clearly, the set consider is an open set in the topology. This open set contains the minimum of the two elements but not the maximum of the two elements. Hence the space is a \(T_0\) space.
• \(X = \mathbb{N}\) and \(\tau\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{n,n+1,\ldots\}\) for any positive integer \(n\). This space is a \(T_0\) space since given any two distinct elements say \(m,n \in \mathbb{N}\), we can consider the set \(\{\max(m,n),\max(m,n)+1,\ldots\}\). Clearly, the set consider is an open set in the topology. This open set contains the maximum of the two elements but not the minimum of the two elements. Hence the space is a \(T_0\) space.
6. Let \(X\) be any infinite set. The countable-closed topology is defined to be the topology having as its closed sets \(X\) and all countable subsets of \(X\). Prove that this is indeed a topology on \(X\).
• Again the same drill of checking the definitions.
• Clearly, \(\emptyset,X \in \tau\).
• Let \(A_{\alpha} \in \tau, \forall \alpha \in \Gamma\) and let \(A = \bigcup_{\alpha \in \Gamma} A_{\alpha}\). This means \(A_{\alpha}^c\) is a closed set \(\forall \alpha \in \Gamma\) i.e. \(A_{\alpha}^c\) is a countable subset of \(X\). This trivially means \(A^c = \bigcap_{\alpha \in \Gamma} A_{\alpha}^c\) is a countable set. (This is so since for instance, the cardinality of the intersection can be atmost the cardinality of one of the \( A_{\alpha}\)'s.) Hence, \(A \in \tau\).
• Consider \(A_1, A_2 \in \tau\) and let \(A = A_1 \bigcap A_2\). This means \(A_1^c,A_2^c \) are countable subsets of \(X\). We get \(A^c = A_1^c \bigcup A_2^c\). But a finite union of countable subsets is again countable which means \(A^c\) is countable and hence \(A \in \tau\). \(\Box\)
7. Let \(\tau_1\) and \(\tau_2\) be two topologies on a set \(X\). Prove each of the following statements.
• If \(\tau_3\) is defined by \(\tau_3 = \tau_1 \cup \tau_2\), then \(\tau_3\) is not necessarily a topology on \(X\).
• Example: Let \(X = \{a,b,c\}\). Let \(\tau_1 = \{\emptyset, \{a\}, X\}\) and \(\tau_2 = \{\emptyset, \{b\}, X\}\). \(\tau_3 = \tau_1 \cup \tau_2\) is not a topology since it doesn't contain \(\{a\} \cup \{b\}\).
• If \(\tau_3\) is defined by \(\tau_3 = \tau_1 \cap \tau_2\), then \(\tau_3\) is a topology on \(X\).
• Again perform the drill
• Clearly, \(\emptyset,X \in \tau_3\) since \(\emptyset,X \in \tau_1\) and \(\emptyset,X \in \tau_2\) as \(\tau_1, \tau_2\) are topologies
• Consider \(A_{\alpha} \in \tau_3 \forall \alpha \in \Gamma\). Since \(\tau_3 = \tau_1 \cap \tau_2\), we have that \(A_{\alpha} \in \tau_1 \forall \alpha \in \Gamma\) and \(A_{\alpha} \in \tau_2 \forall \alpha \in \Gamma\). Let \(A = \bigcup_{\alpha \in \Gamma} A_{\alpha}\). Since \(\tau_1\) and \(\tau_2\) are topologies, we now have that \(A \in \tau_1\) and \(A \in \tau_2\). This means that \(A \in \tau_3\). Hence, \(\tau_3\) is closed under arbitrary unions.
• Consider \(A_1, A_2 \in \tau_3\). Since \(\tau_3 = \tau_1 \cap \tau_2\), we have that \(A_1,A_2 \in \tau_1\) and \(A_1,A_2 \in \tau_2\). Let \(A = A_1 \bigcap A_2\). Since \(\tau_1,\tau_2\) are topologies, we get \(A \in \tau_1\) and \(A \in \tau_2\). Hence, \(A \in \tau_3\). Hence, \(\tau_3\) is closed under finite intersections.
• If \(X,\tau_1\) and \(X,\tau_2\) are \(T_1\)-spaces, and let \(\tau_3 = \tau_1 \cap \tau_2\) then \(X,\tau_3\) is also a \(T_1\)-space.
• \(T_1\)-space is one in which the topology has every singleton set as a closed set. This means \(\forall x \in X\), we have \(X \backslash \{x\} \in \tau_1\) and \(X \backslash \{x\} \in \tau_2\). Since \(\tau_3 = \tau_1 \cap \tau_2\), we have \(X \backslash \{x\} \in \tau_3\), \(\forall x \in X\). Hence, every singleton set as a closed set under the topology \(\tau_3\). Hence, the space \(X, \tau_3\) is a \(T_1\)-space.
• If \(X,\tau_1\) and \(X,\tau_2\) are \(T_0\)-spaces, and let \(\tau_3 = \tau_1 \cap \tau_2\) then \(X,\tau_3\) is not necessarily a \(T_0\)-space.
• Example: Let \(X = \{a,b,c\}\) and let \(\tau_1 = \{\emptyset, \{a\}, \{b\}, \{a,b\}, X\}\) and \(\tau_2 = \{\emptyset, \{b\}, \{c\}, \{b,c\}, X\}\). It is not hard to check that \(\tau_1\) and \(\tau_2\) induce a \(T_0\)-space. However, \(\tau_3 = \{\emptyset, \{b\}, X\}\) does not induce a \(T_0\)-space.
• If for each \(\alpha \in \Gamma\), each \(\tau_{\alpha}\) is a topology on the set \(X\) then \(\tau = \bigcap_{\alpha \in \Gamma} \tau_{\alpha}\) is a topology on \(X\). (Combined the last two questions in the textbook.)
• Again the drill.
• Clearly, \(\emptyset,X \in \tau_{\alpha}, \forall \alpha \in \Gamma\) and hence \(\emptyset,X \in \bigcap_{\alpha \in \Gamma} \tau_{\alpha}\).
• Consider \(A_{\beta} \in \tau\), \(\forall \beta \in \Delta\). Let \(A = \bigcup_{\beta \in \Delta} A_{\beta}\). Since \(\tau= \bigcap_{\alpha \in \Gamma} \tau_{\alpha}\), we have that \(A_{\beta} \in \tau_{\alpha}\), \(\forall \beta \in \Delta\) and \(\forall \alpha \in \Gamma \). Since each of the \(\tau_{\alpha}\) are topologies, we get that \(A \in \tau_{\alpha}\), \(\forall \alpha \in \Gamma \) Hence, \(A \in \tau\). Hence, \(\tau\) is closed under arbitrary unions.
• Consider \(A_1,A_2 \in \tau\). Let \(A = A_1 \bigcap A_2\). Since \(\tau= \bigcap_{\alpha \in \Gamma} \tau_{\alpha}\), we have that \(A_1,A_2 \in \tau_{\alpha}, \forall \alpha \in \Gamma\). Since every \(\tau_{\alpha}\) is a topology, we get that \(A \in \tau_{\alpha, \forall \alpha \in \Gamma}\). Hence, \(A \in \tau\). Hence, \(\tau\) is closed under finite intersections.