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.