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.