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.