Topology without Tears - 2.2 Exercises

Exercises:

1. In this exercise, you will prove that the disc \(\{(x,y): x^2 + y^2 < 1\}\) is an open subset of \(\mathbb{R}^2\), and then that every open disc in the plane is an open set.
• Let \((a,b)\) be any point in the disc \(D = \{(x,y):x^2+y^2<1\}\). Put \(r = \sqrt{a^2 + b^2}\). Let \(R_{(a,b)}\) be the open rectangle with vertices at the points \(\left(a \pm \frac{1-r}{8}, b \pm \frac{1-r}{8} \right)\). Verify that \(R_{(a,b)} \subset D\).

• All we need to show is that all the four vertices lies inside the disc, for which we need to show that the distance from the origin to each of the four vertices is less than unity. To show that all we need to show is that the distance of the farthest vertex from the origin is less than unity. The distance of the farthest vertex from the origin is given by \(d = r+ \sqrt{2} \frac{1-r}{8} = \frac{\sqrt{2}}{8} + r \left(1 - \frac{\sqrt{2}}{8} \right) \). Since \(0 \leq r \lt 1\), we get \(\frac{\sqrt{2}}{8} \leq d \lt 1\). Hence, we get that \(R_{(a,b)} \subset D\).
• Using the previous part show that \(D = \bigcup_{(a,b) \in D} R_{(a,b)}\).
• Consider any point \((a,b) \in D\). Clearly, \((a,b) \in R_{(a,b)}\). Hence, any point in the disc is contained in an open rectangle centered about that point. Hence, \(D \subseteq \bigcup_{(a,b) \in D} R_{(a,b)}\). From the previous part, we have that for any \((a,b)\) we have \(R_{(a,b)} \subseteq D\). Hence, \(D \supseteq \bigcup_{(a,b) \in D} R_{(a,b)}\). Combining the two, we get \(D = \bigcup_{(a,b) \in D} R_{(a,b)}\).
• Deduce from above that \(D\) is an open set in \(\mathbb{R}^2\).
• Every open rectangle is an open set. Hence, any arbitrary union of open sets is again an open set. From the previous part, we have that \(D\) is an arbitrary union of open rectangles. Hence, \(D\) is an open set.
• Show that every disc \(\{(x,y): (x-a)^2 + (y-b)^2 \lt c^2, a,b,c \in \mathbb{R}\}\) is open in \(\mathbb{R}^2\).
• Let \((m,n)\) be any point in the disc \(D = \{(x,y): (x-a)^2 + (y-b)^2 \lt c^2\}\). Put \(r = \sqrt{(m-a)^2 + (n-b)^2}\). Let \(R_{(a,b)}\) be the open rectangle with vertices at the points \((a + m \pm \frac{c - r}{8}, b + n \pm \frac{c-r}{8})\). Now the same arguments as the above three gives us that the disc \(D = \{(x,y): (x-a)^2 + (y-b)^2 \lt c^2\}\) is open in \(\mathbb{R}^2\).
2. In this exercise you will show that the collection of all open discs in \(\mathbb{R}^2\) is a basis for a topology on \(\mathbb{R}^2\). [Later we shall see that this is the euclidean topology.]
• Let \(D_1\) and \(D_2\) be any open discs in \(\mathbb{R}^2\) with \(D_1 \cap D_2 \neq \emptyset\). If \((a,b)\) is any point in \(D_1 \cap D_2\), show that there exists an open disc \(D_{(a,b)}\) with centre \((a,b)\) such that \(D_{(a,b)} \subseteq D_1 \cap D_2\).

• We are given that \(D_1 \cap D_2 \neq \emptyset\). Let the center and radius of \(D_1\) be \((a_1,b_1)\) and \(r_1\) respectively. Let the center and radius of \(D_2\) be \((a_2,b_2)\) and \(r_2\) respectively. Since, we have \(D_1 \cap D_2 \neq \emptyset\), we have \(r_1 + r_2 \gt \sqrt{(a_1-a_2)^2 + (b_1-b_2)^2}\). Let \(r = \min(r_1-\sqrt{(a-a_1)^2+(b-b_1)^2},r_2-\sqrt{(a-a_2)^2 + (b-b_2)^2})\). Consider the disc centered at \((a,b)\) with radius \(r\). This disc lies completely inside \(D_1 \cap D_2\).
• Show that \[D_1 \cap D_2 = \bigcup_{(a,b) \in D_1 \cap D_2} D_{(a,b)}.\]
• Consider any \((x,y) \in D_1 \cap D_2\). Note that \((x,y) \in D_{(x,y)}\). \((x,y)\) is one such \((a,b)\) in the union. Hence \((x,y) \in D_{(x,y)} \subseteq \bigcup_{(a,b) \in D_1 \cap D_2} D_{(a,b)}\). Hence, we get \(D_1 \cap D_2 \subseteq \bigcup_{(a,b) \in D_1 \cap D_2} D_{(a,b)}\). Now, for any \((x,y) \in D_1 \cap D_2\), we have \(D_{(x,y)} \subseteq D_1 \cap D_2\). Hence, we get \(\bigcup_{(a,b) \in D_1 \cap D_2} D_{(a,b)} \subseteq D_1 \cap D_2\).
• Using the above and the proposition proved earlier, prove that the collection of all open discs in \(\mathbb{R}^2\) is a basis for a topology on \(\mathbb{R}^2\).
• First note that we have \(\mathbb{R}^2 = \bigcup D_{(a,b)}(n)\) where \(D_{(a,b)}(n)\) denotes an open unit disc of radius \(n\) centered at \((a,b)\). The proof is trivial since any \(D_{(a,b)}(n) \subseteq \mathbb{R}^2\) and also for any point \((x,y) \in \mathbb{R}^2\), we have \(n \in \mathbb{N}\) such that \(n \gt \sqrt{x^2 + y^2}\) and hence \((x,y) \in D_{(0,0)}(n)\). Next note that from the part above, we have that intersection of any two discs is again a union of open discs. Hence, by the proposition proved earlier we have that the collection of all open discs in \(\mathbb{R}^2\) is a basis for a topology on \(\mathbb{R}^2\).
3. Let \(\mathcal{B}\) be the collection of all open intervals \((a,b)\) in \(\mathbb{R}\) with \(a \lt b\) and \(a\) and \(b\) rational numbers. Prove that \(\mathcal{B}\) is a basis for the euclidean topology on \(\mathbb{R}\).
• We shall prove that if \(S \subseteq \mathbb{R}\) is an open set then given any \(x \in S\) we can find an open interval \((a_q,b_q) \subseteq S\) where \(a_q,b_q \in \mathbb{Q}\). Since \(S\) is an open set, we know that there exists \(a,b \in \mathbb{R}\) such that \(x \in (a,b) \subseteq S\). Further, since the rationals are dense in \(\mathbb{R}\), given any \(r \in \mathbb{R}\), there exists a sequence of rationals monotonously converging to \(r\). For instance, there exists a sequence of monotonously increasing \(b_n \in \mathbb{Q}\) such that \(\lim_{n \rightarrow \infty}b_n = b\). Similarly, there exists a sequence of monotonously decreasing \(a_n \in \mathbb{Q}\) such that \(\lim_{n \rightarrow \infty}a_n = a\). This means that \((a,b) = \bigcup_{n \rightarrow \infty} (a_n,b_n)\). Hence, if \(x \in (a,b)\), then \(x \in (a_n,b_n)\) for some \(n\) where \(a_n,b_n \in \mathbb{Q}\). Hence, given any open set \(S\), for every \(x \in S\), there exists an open interval with rational end points containing \(x\) lying within \(S\). Hence, the set of open intervals with rational end points generate the same topology as the euclidian topology.
4. A topological space \((X,\tau)\) is said to satisfy the second axiom of countability or to be second countable if there exists a basis \(\mathcal{B}\) for \(\tau\), where \(\mathcal{B}\) consists of only a countable number of sets.
• Using the previous exercise show that \(\mathbb{R}\) satisfies the second axiom of countability.
• We have \(\mathcal{B} = \{(a_q,b_q): a_q,b_q \in \mathbb{Q}\}\)  is a basis for the euclidean topology. We have the set of rational numbers to be countable. The set \(\mathcal{B}\) can be written as \(\mathcal{B} = \bigcup_{a_q \lt b_q; a_q \in \mathbb{Q}} \bigcup_{b_q \in \mathbb{Q}} (a_q,b_q)\) which is again a countable set.
•  Prove that the discrete topology on an uncountable set does not satisfy the second axiom of countability.
• Let \(\mathcal{B}\) be a basis for the discrete topology. For every \(x \in X\), we have \(\{x\}\) to be an open set in the discrete topology. This means that we have \(\{x\}\) to be a union of elements in \(\mathcal{B}\). This means that the singleton sets should be in the basis \(\mathcal{B}\). Hence, we have \(\{\{x\} : x \in X\} \subseteq \mathcal{B} \). Since \(X\) is uncountable, we have \( \{\{x\} : x \in X\}\) to be an uncountable set. Hence, we have that \(\mathcal{B}\) to be an uncountable set. Hence, \((X,\tau)\) satisfies the second axiom of countability.
• Prove that \(\mathbb{R}^n\) satisfies the second axiom of countability, for each positive integer \(n\).
• We proceed by induction. Let \(P(n)\) be the statement that \(\mathbb{R}^n\) satisfies the second axiom of countability. Let \(S = \{n \in \mathbb{N}: P(n) \text{ is true}\}\). From the first part of this problem, we have \(1 \in S\) i.e there exists a countable basis \(\mathcal{B}\) for the euclidean topology on \(\mathbb{R}\). Assume that \(k \in S\). Note that \(\mathbb{R}^{k+1} = \mathbb{R}^k \times \mathbb{R}\). By induction hypothesis, we have that there exists a countable basis \(\mathcal{B}^k\) for the euclidean topology on \(\mathbb{R}^k\). From the question \(6\), which is proved later we have that \(\mathcal{B}^k \times \mathcal{B}\) is a basis for \(\mathbb{R}^k \times \mathbb{R} = \mathbb{R}^{k+1}\). Product of two countable sets is again a countable set. Hence, \(\mathcal{B}^k \times \mathcal{B}\) is a countable basis for \(\mathbb{R}^k \times \mathbb{R} = \mathbb{R}^{k+1}\). Hence, \(\mathbb{R}^n\) satisfies the second axiom of countability, for each positive integer \(n\).
• Let \((X,\tau)\) be the set of all integers with the finite-closed topology. Does the space \((X,\tau)\) satisfy the second axiom of countability?
• Let \(\tau^c = \{A \subseteq X: X \backslash A \in \tau\}\) i.e. \(\tau^c\) contains all the closed sets induced by the topology \(\tau\). In this case, we have \(\tau^c = \{A \subseteq X : A \text{ is finite}\}\). Note that \(\tau\) is equivalent to \(\tau^c\) since there is a clear bijection from \(\tau\) to \(\tau^c\) as \(A \in \tau\) iff \(X \backslash A \in \tau^c\). We shall prove that \(\tau^c\) is a countable set. This would mean that \(\tau\) is a countable set and since any basis \(\mathcal{B} \subseteq \tau\) we would have proved that the \((X,\tau)\), with the finite-closed topology, satisfies the second axiom of countability. Since \(X\) is a countable set, list the element of \(X\) as \(\{x_0,x_1,\ldots,x_n,\ldots\}\). Any finite subset \(B\) of \(X\) is of the form \(B = \{x_{k_0},x_{k_1},\ldots,x_{k_n}\}\) where \(n \in \mathbb{N}\) and \(k_i \in \mathbb{N}\) for \(i \in \{0,1,\ldots,n\}\). Let \(\displaystyle f(B) = \sum_{l=0}^{n} 2^{k_l}\). It is not hard to see that \(f: \tau^c \rightarrow \mathbb{N}\) is a bijection. Hence, \(\tau^c\) is a countable set. This means that the topology, \(\tau\), is also countable and hence any basis is also a countable set. Hence, \((X,\tau)\), with the finite-closed topology, satisfies the second axiom of countability.
5. Prove the following statements.
• Let \(m\) and \(c\) be real numbers, with \(m \neq 0\). Then the line \(L = \{(x,y): y = mx+c\}\) is a closed subset of \(\mathbb{R}^2\).

• Consider \(L^c = \mathbb{R}^2 \backslash L\). We shall prove that \(L^c\) is an open set. Consider \((a,b) \in L^c\). We shall prove that there is a rectangle \(R \subseteq L^c\) such that \((a,b) \in R\). Let \(d\) denote the distance of the point \((a,b)\) from the line \(L\). We have \(d = \frac{b-am-c}{\sqrt{1+m^2}} \gt 0\). Consider the open rectangle \(R\) with vertices \((a \pm \frac{d}{2}, b \pm \frac{d}{2})\). We get \((a,b) \in R\) and \(R \subseteq L^c\). Hence, \(L^c\) is an open set. Hence, \(L\) is a closed set.
• Let \(\mathbb{S}^1\) be the unit circle given by \(\mathbb{S}^1 = \{(x,y) \in \mathbb{R}^2: x^2 + y^2 = 1\}\). Then \(\mathbb{S}^1\) is a closed subset of \(\mathbb{R}^2\).
• Consider \(\mathbb{T} = \mathbb{R}^2 \backslash \mathbb{S}^1\). We shall prove that \(\mathbb{T}\) is open. Consider any \((a,b) \in \mathbb{T}\). Let \(r = \text{abs}(1-\sqrt{a^2+b^2})\). Consider the open rectangle \(R\) with vertices at \(( a \pm \frac{r}{2}, b \pm \frac{r}{2})\). Clearly, we have \((a,b) \in R \subseteq \mathbb{T}\). This is true for any \((a,b) \in \mathbb{T}\). Hence, \(\mathbb{T}\) is an open set. Hence, \(\mathbb{S}^1\) is a closed set of \(\mathbb{R}^2\).
• Let \(\mathbb{S}^n\) be the unit \(n\)-sphere given by \[\mathbb{S}^n = \{(x_1,x_2,\ldots,x_n,x_{n+1}) \in \mathbb{R}^{n+1}: x_1^2 + x_2^2 + \cdots + x_{n+1}^2 = 1\}.\] Then \(\mathbb{S}^n\) is a closed subset of \(\mathbb{R}^{n+1}\).
• Consider \(\mathbb{T} = \mathbb{R}^{n+1} \backslash \mathbb{S}^n\). We shall prove that \(\mathbb{T}\) is open. Consider any \((a_1,a_2,\ldots,a_{n+1}) \in \mathbb{T}\). Let \(r = \text{abs} \left(1 - \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2 + a_{n+1}^2} \right)\). Consider the open rectangle \(R\) with vertices at \( \left(a_1 \pm \frac{r}{n+1}, a_2 \pm \frac{r}{n+1}, \ldots, a_{n} \pm \frac{r}{n+1}, a_{n+1} \pm \frac{r}{n+1} \right)\). Clearly, we have \((a_1,a_2,\ldots,a_n,a_{n+1}) \in R \subseteq \mathbb{T}\). Hence, \(\mathbb{T}\) is an open set. Hence, \(\mathbb{S}^n\) is a closed subset of \(\mathbb{R}^{n+1}\).
• Let \(B^n\) be the closed unit \(n\)-ball given by \[B^n = \{(x_1,x_2,\ldots,x_{n}): x_1^2 + x_2^2 + \cdots + x_n^2 \leq 1\}.\] Then \(B^n\) is a closed subset of \(\mathbb{R}^n\).
• Consider \(\mathbb{T} = \mathbb{R}^n \backslash B^n\). We shall prove that \(\mathbb{T}\) is open. Consider any \((a_1,a_2,\ldots,a_n) \in \mathbb{T}\). Let \(r = \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} - 1\). Consider the open rectangle \(R\) with vertices at \( \left(a_1 \pm \frac{r}{n+1}, a_2 \pm \frac{r}{n+1}, \ldots, a_{n} \pm \frac{r}{n+1}, a_{n+1} \pm \frac{r}{n+1} \right)\). Clearly, we have \((a_1,a_2,\ldots,a_n,a_{n+1}) \in R \subseteq \mathbb{T}\). Hence, \(\mathbb{T}\) is an open set. Hence, \(B^n\) is a closed subset of \(\mathbb{R}^{n}\).
• The curve \(C = \{(x,y) \in \mathbb{R}^2 : xy=1\}\) is a closed subset of \(\mathbb{R}^2\).
• Consider \(\mathbb{T} = \mathbb{R}^2 \backslash C\). We shall prove that \(\mathbb{T}\) is open. Consider any \((a,b) \in \mathbb{T}\). Let \(r\) be the minimum distance from the point \((a,b)\) to the curve \(C\). Consider the open rectangle \(R\) with vertices at \(( a \pm \frac{r}{2}, b \pm \frac{r}{2})\). Clearly, we have \((a,b) \in R \subseteq \mathbb{T}\). This is true for any \((a,b) \in \mathbb{T}\). Hence, \(\mathbb{T}\) is an open set. Hence, \(C\) is a closed set of \(\mathbb{R}^2\).
6. Let \(\mathcal{B}_1\) be a basis for the topology \(\tau_1\) on a set \(X\) and \(\mathcal{B}_2\) be a basis for the topology \(\tau_2\) on a set \(Y\). The set \(X \times Y\) consists of all ordered pairs \((x,y)\), \(x \in X\) and \(y \in Y\). Let \(\mathcal{B}\) be the collection of subsets of \(X \times Y\) consisting of all the sets \(B_1 \times B_2\) where \(B_1 \in \mathcal{B}_1\) and \(B_2 \in \mathcal{B}_2\). Prove that \(\mathcal{B}\) is a basis for a topology on \(X \times Y\). The topology so defined is called the product topology on \(X \times Y\).
• All we need to do is to check the definitions of a basis. We are given that \(\mathcal{B}_1\) and \(\mathcal{B}_2\) are a basis for \(\tau_1\) and \(\tau_2\) respectively. Hence, we have \(\cup_{B \in \mathcal{B}_1} B = X\) and \(\cup_{B \in \mathcal{B}_2} B = Y\). Also, if we have \(C,D \in \mathcal{B}_1\), then \(C \cap D \in \mathcal{B}_1\). Similarly, if we have \(C,D \in \mathcal{B}_2\), then \(C \cap D \in \mathcal{B}_2\). Now consider any element in \(\mathcal{B}_1 \times \mathcal{B}_2\). It is of the form \(B_1 \times B_2\) where \(B_1 \in \mathcal{B}_1 \) and \(B_2 \in \mathcal{B}_2 \). Now the union over all elements in \(\mathcal{B}_1 \times \mathcal{B}_2\) can be written as \(\cup_{B_1 \in \mathcal{B}_1 , B_2 \in \mathcal{B}_2} B_1 \times B_2\). We then have \(\cup_{B_1 \in \mathcal{B}_1 , B_2 \in \mathcal{B}_2} B_1 \times B_2 = \cup_{B_1 \in \mathcal{B}_1} \left( B_1 \times \cup_{B_2 \in \mathcal{B}_2} B_2 \right) = \cup_{B_1 \in \mathcal{B}_1} B_1 \times Y = \cup_{B_1 \in \mathcal{B}_1} \left( B_1 \right) \times Y = X \times Y\). Also, if we have \(A,B \in \mathcal{B}\), then \(A = A_1 \times A_2\) and \(B= B_1 \times B_2\) where \(A_1,B_1 \in \mathcal{B}_1\) and \(A_2,B_2 \in \mathcal{B}_2\). Then \(A \cap B = \left( \right)\).