Topology without Tears - 1.2 Exercises

Exercises:

1. List all \(64\) subsets of the set \(X\) where \(X = \{a,b,c,d,e,f\}\) and the topology \(\tau = \{X,\emptyset,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e,f\}\}\). Write down, next to each set, whether it is clopen, (or) neither open nor closed (or) open but not closed (or) closed but not open with respect to the topology \( \tau\)
• The clopen sets are \(X, \emptyset, \{a\},\{b,c,d,e,f\}\)
• The sets that are open but not closed are \(\{c,d\},\{a,c,d\}\)
• The sets that are closed but not open are \(\{a,b,e,f\},\{b,e,f\}\)
• The rest of the sets are neither open nor closed
2. Let \((X,\tau)\) be a topological space with the property that every subset is closed. Prove that it is a discrete space.
• Consider any subset \(A \subseteq X\). Consider \(A^c\). Since \(A^c\) is also a subset of \(X\) and from the problem we get that \(A^c\) is closed. Hence, \(A = {(A^c)}^c\) is open. Hence, \(A \in \tau\), \(\forall A \subseteq X\). Hence, \((X,\tau)\) is a discrete topological space.
3. Observe that if \((X,\tau)\) is a discrete space (or) an indiscrete space, then every open set is clopen. Find a topology \(\tau\) on a set \(X=\{a,b,c,d\}\) which is not discrete and is not indiscrete but has the property that every set is clopen.
• Consider \(\tau = \{\emptyset,\{a\},\{b,c,d\},X\}\). Clearly, \(\tau\) is a topology and every set in \(\tau\) is clopen.
4. Let \(X\) be an infinite set. If \(\tau\) is a topology on \(X\) such that every infinite subset of \(X\) is closed, prove that \(\tau\) is the discrete topology.
• Consider any element \(x \in X\). We shall prove that \(\{x\} \in \tau\). Let \(A_x = X \backslash \{x\}\). Clearly, \(A_x\) is an infinite set. (If not, then \(X = A_x \bigcup \{x\} \implies X \text{ is finite}\)). Since it is given that every infinite set is closed, we have \(A_x\) is closed and hence \(\{x\}\) is open. Hence, every singleton set is open. Hence by a proposition proved earlier we get that \(\tau\) is the discrete topology
5. Let \(X\) be an infinite set and \(\tau\) a topology on \(X\) with the property that the only infinite subset of \(X\) which is open is \(X\) itself. Is \((X,\tau)\) necessarily an indiscrete space?
• No. For instance, consider any \(x \in X\). Let \(\tau_x = \{\emptyset,\{x\},X\}\). Clearly, \(\tau_x\) is a topology on \(X\). But \((X,\tau)\) is an indiscrete space.
6. Let \(\tau\) be a topology on a set \(X\) such that \(\tau\) consists of precisely four sets; i.e. \(\tau = \{X,\emptyset,A,B\}\), where \(A\) and \(B\) are non-empty distinct proper subsets of \(X\). [\(S\) is a proper subset of \(X\) means \(S \subseteq X\) and \(S \neq X\). It is denoted by \(S \subset X\)]. Prove that \(A\) and \(B\) must satisfy exactly one of the following conditions:
• \(B = X \backslash A \)
• \(A \subset B\)
• \(B \subset A\)
• By property \(2\), in the definition of a topology, we need \(A \bigcup B \in \tau\). Since \(A\) and \(B\) are non-empty, we need \(A \bigcup B = A\) (or) \(A \bigcup B = B\) (or) \(A \bigcup B = X\). 
• If  \(A \bigcup B = A\), then we get \(B \subset A\). (since \(A\) and \(B\) are distinct, \(B \neq A\)).
• If  \(A \bigcup B = B\), then we get \(A \subset B\). (since \(A\) and \(B\) are distinct, \(A \neq B\)).
• Hence, the only other option is \(A \bigcup B = X\). Now make use of property \(3\) in the definition of topology to get that \(A \bigcap B\) can be either \(\emptyset\) (or) \(A\) (or) \(B\). (It cannot be \(X\) since \(A\) and \(B\) are both proper subsets of \(X\))
• If \(A \bigcap B = A\), then \(A \subset B\) and hence \(A \bigcup B = B \neq X\)
• If \(A \bigcap B = B\), then \(B \subset A\) and hence \(A \bigcup B = A \neq X\)
• If \(A \bigcap B = \emptyset\), we also need \(A \bigcup B = X\) which gives us that \(B = X \backslash A\)
• Using the above list all topologies on \(X = \{1,2,3,4\}\) which consist of exactly four sets
• \(\tau = \{\emptyset,\{1\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,2\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,3\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,4\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,2\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{2,3\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{2,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{2,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{4,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{2,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{3,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{1,2\},\{3,4\},X \} \)
• \(\tau = \{\emptyset,\{1,2\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{1,2\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{1,3\},\{2,4\},X \} \)
• \(\tau = \{\emptyset,\{1,3\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{1,3\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{1,4\},\{2,3\},X \} \)
• \(\tau = \{\emptyset,\{1,4\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{1,4\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2,3\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{2,3\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2,4\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{2,4\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{3,4\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{3,4\},\{2,3,4\},X \} \)