Topology without Tears - 1.3 Definitions and Propositions

Definition:

Let \(X\) be any non-empty set. A topology \(\tau\) on \(X\) is called the finite-closed topology (or) the co-finite topology if the closed subsets of \(X\) are \(X\) and all finite subsets of \(X\); i.e. the open sets are \(\emptyset\) and all subsets of \(X\) which have finite complements i.e. \(\tau = \{A \subseteq X: A = \emptyset \text{ (or) } X \backslash A \text{ is finite}\}\)

Proposition:

Let \(\tau\) be the finite-closed topology on a set \(X\). If \(X\) has at-least \(3\) distinct clopen subsets, prove that \(X\) is a finite set.

Proof:

We are given there are at-least \(3\) clopen subsets. Note that in any topology \(\emptyset\) and \(X\) are always clopen. Hence there exists \(A \in \tau \) with \(\emptyset \subset A \subset X\) so that \(A^c \in \tau\). This means that since \(A\) is also closed and since the topology is a co-finite topology \(A\) is a finite set and similarly, since \(A^c\) is also closed \(A^c\) is also a finite set. But \(X = A \bigcup A^c \implies X\) is a finite set. Hence proved. \(\Box\)

Definition:

Let \(f\) be a function from a set \(X\) to a set \(Y\).

1. The function \(f\) is said to be one-to-one (or) injective if \(f(x_1) = f(x_2) \implies x_1 = x_2 \forall x_1,x_2 \in X\);
2. The function \(f\) is said to be onto (or) surjective if for each \(y \in Y\) there exists an \(x \in X\) such that \(f(x) = y\);
3. The function \(f\) is said to be bijective if it is both injective and surjective.

Definition:

Let \(f\) be a function from a set \(X\) into a set \(Y\). The function \(f\) is said to have an inverse if there exists a function \(g\) from \(Y\) to \(X\) such that \(g(f(x)) = x\) and \(f(g(y)) = y\) for all \(y \in Y\). The function \(g\) is called an inverse function of \(f\).

Proposition:

Let \(f\) be a function from a set \(X\) into a set \(Y\).

1. The function \(f\) has an inverse if and only if \(f\) is bijective.
2. Let \(g_1\) and \(g_2\) be functions from \(Y\) into \(X\). If \(g_1\) and \(g_2\) are both inverse functions of \(f\), then \(g_1 = g_2\); that is, \(g_1(y) = g_2(y)\), for all \(y \in Y\).
3. Let \(g\) be a function from \(Y\) into \(X\). Then \(g\) is an inverse function of \(f\) if and only if \(f\) is an inverse function of \(g\).

Proof:

1. We shall first prove that if \(f\) has an inverse then \(f\) is bijective and then the other way around.
• If \(f\) has an inverse, then it is bijective. Note that if \(f\) from \(X\) to \(Y\) has an inverse, then there exists a function \(g\) from \(Y\) to \(X\) such that \(g(f(x)) = x\) and \(f(g(y)) = y\) for all \(y \in Y\).
• We shall prove that \(f\) is injective. Consider \(x_1,x_2 \in X\). Let \(f(x_1) = f(x_2) = y\). Since \(g\) is a function from \(Y\) to \(X\), we get \(g(y) = g(f(x_1)) = g(f(x_2))\). From the definition of the inverse function, we have \(g(f(x_1)) = x_1\) and \(g(f(x_2)) = x_2 \). But we have \(g(f(x_1)) = g(f(x_2))\). Hence, we get \(x_1 = x_2\). Hence, \(f\) is injective.
• We shall now prove that \(f\) is surjective. Consider any \(y \in Y\). Let \(x = g(y)\). We now get \(f(x) = f(g(y)) = y\). Hence, \(\exists x \in X\) such that \(f(x) = y\). Hence for each \(y \in Y\) there exists \(x \in X\) such that \(f(x)=y\). In fact, we even know what \(x\) is. \(x\) is given by \(g(y)\). Hence, the function is surjective.
• If \(f\) is bijective, then \(f\) has an inverse.
• Since \(f\) is bijective, given any \(y \in Y\), we can find a unique \(x \in X\) such that \(y = f(x)\). Since the \(x\) is unique, map \(y\) to \(x\). Let this mapping be \(g\). This defines a well-defined function from \(Y\) to \(X\). It is well-defined since it is defined for all \(y \in Y\) (since \(f\) is surjective)  and also \(g(y)\) is a unique number in \(X\) (since \(f\) is injective). Hence, we get \(g(f(x)) = g(y) = x\) and \(f(g(y)) = f(x) = y\). Hence, \(g\) is an inverse for \(f\).
2. We will now prove that the inverse is unique i.e. if \(g_1\) and \(g_2\) are both inverse functions of \(f\), then \(g_1 = g_2\); that is, \(g_1(y) = g_2(y)\), for all \(y \in Y\).
• Consider \(y \in Y\). Since \(f\) is bijective, there exists a unique \(x \in X\) such that \(f(x) = y\). Since \(g_1\) and \(g_2\) are both inverses, we get \(g_1(y) = g_1(f(x)) = x\) and \(g_2(y) = g_2(f(x)) = x\). Hence, we get that \(\forall y \in Y\), we have \(g_1(y) = g_2(y)\). Hence, \(g_1 = g_2\).
3. We shall now prove the last part, namely, \(g\) is an inverse function of \(f\) if and only if \(f\) is an inverse function of \(g\)
• We shall first prove that if \(g\) is an inverse function of \(f\) then \(f\) is an inverse function of \(g\). We are given that there exists a function \(g: Y \rightarrow X\) such that \(g(f(x)) = x\) and \(f(g(y)) = y\). Note that from this it immediately follows that \(f: X \rightarrow Y\) such that \(f(g(y)) = y\) and \(g(f(x)) = x\). Hence \(f\) is the inverse of \(g\).
• The argument to prove the other way around, i.e. if \(f\) is an inverse function of \(g\) then \(g\) is an inverse function of \(f\) is the same as the argument for the above part. \(\Box\)

Definition:

Let \(f\) be a function from a set \(X\) into a set \(Y\). If \(S\) is any subset of \(Y\), then the set \(f^{-1}(S)\) is defined by \[f^{-1}(S) = \{x: x \in X \text{ and }f(x) \in S \}.\] The subset \(f^{-1}(S)\) of \(X\) is said to be the inverse image of \(S\)

Proposition:

Let \(T,\tau\) be a topological space and \(X\) be a non-empty set. Further, let \(f\) be a function from \(X\) into \(Y\). Let \(\tau_x =\{f^{-1}(S): S \in \tau \}\). Prove that \(\tau_x\) is a topology on \(X\).

Proof:

As usual, we need to check whether \(\tau_x\) satisfies the three properties

1. Note that \(f^{-1}(\emptyset ) = \emptyset\) and \(f^{-1}(Y) = X\). Since \(\emptyset, Y \in \tau_Y\), we get \(\emptyset, X \in \tau_X\)
2. Let \(A_{\alpha} \in \tau_x\), \(\forall \alpha \in \Gamma\). This means that there exists \(S_{\alpha} \in \tau\) such that \(f^{-1}(S_{\alpha}) = A_{\alpha}\). Let \(S = \bigcup_{\alpha \in \Gamma} S_{\alpha} \). We claim that \(f^{-1}(S) = \bigcup_{\alpha \in \Gamma} f^{-1}(S_{\alpha})\). The proof for the claim can be seen below. Now, \(S \in \tau\) since \(\tau\) is a topology. This means \(f^{-1}(S) \in \tau_x\) from the definition of \(\tau_x\). But \(f^{-1}(S) = \bigcup_{\alpha \in \Gamma} f^{-1}(S_{\alpha}) = \bigcup_{\alpha \in \Gamma} A_{\alpha}\). Hence, we get \(\bigcup_{\alpha \in \Gamma} A_{\alpha} \in \tau_x\). Hence, \(\tau_x\) is closed under arbitrary unions.
• Claim: \(f^{-1}(S) = \bigcup_{\alpha \in \Gamma} f^{-1}(S_{\alpha})\) where \(S = \bigcup_{\alpha \in \Gamma} S_{\alpha}\)
• Proof: 
• We will first show that \(f^{-1}(S) \subseteq \bigcup_{\alpha \in \Gamma} f^{-1}(S_{\alpha})\). Let \(x \in f^{-1}(S)\). This means \(f(x) \in S \implies f(x) \in \bigcup_{\alpha \in \Gamma} S_{\alpha} \implies f(x) \in S_{\alpha}\) for some \(\alpha \in \Gamma\). Hence, we get \(x \in f^{-1}(S_{\alpha})\) for some \(\alpha \in \Gamma\). Hence, \(f^{-1}(S) \subseteq \bigcup_{\alpha \in \Gamma} f^{-1}(S_{\alpha})\).
• We shall now show show that \(f^{-1}(S) \supseteq \bigcup_{\alpha \in \Gamma} f^{-1}(S_{\alpha})\). Let \(x \in \bigcup_{\alpha \in \Gamma} f^{-1}(S_{\alpha})\). This means \(x \in f^{-1}(S_{\alpha}) \) for some \(\alpha \in \Gamma\) which implies \(f(x) \in S_{\alpha}\) for some \(\alpha \in \Gamma\) which implies \(f(x) \in \bigcup_{\alpha \in \Gamma} S_{\alpha} = S \implies x \in f^{-1}(S)\).
3. We now need to prove that if \(A_1, A_2 \in \tau_x\) then so is \(A_1 \cap A_2\). Since  \(A_1, A_2 \in \tau_x\) we have \(S_1, S_2 \in \tau\) such that \(f^{-1}(S_1) = A_1\) and \(f^{-1}(S_2) = A_2\). Since \(S_1,S_2 \in \tau\) and since \(\tau\) is a topology, we get \(S = S_1 \bigcap S_2 \in \tau\). We claim that \(f^{-1}(S) = f^{-1}(S_1) \bigcap f^{-1}(S_2)\). The proof for the claim can be seen below. Now, \(S \in \tau\) since \(\tau\) is a topology. This means \(f^{-1}(S) \in \tau_x\) from the definition of \(\tau_x\). But \(f^{-1}(S) = f^{-1}(S_1) \bigcap f^{-1}(S_2) = A_1 \bigcap A_2\). Hence, we get \(A_1 \bigcap A_2 \in \tau_x\). Hence, \(\tau_x\) is closed under intersections.
• Claim: \(f^{-1}(S) = f^{-1}(S_1) \bigcap f^{-1}(S_2)\) where \(S = S_1 \bigcap S_2\)
• Proof:
• We will first show that \(f^{-1}(S) \subseteq f^{-1}(S_1) \bigcap f^{-1}(S_2)\). Let \(x \in f^{-1}(S)\). This means \(f(x) \in S \implies f(x) \in S_1 \bigcap S_2 \implies f(x) \in S_1 \text{ and } f(x) \in S_1\). Hence, we get \(x \in f^{-1}(S_1)\) and \(x \in f^{-1}(S_2)\). Hence, \(x \in f^{-1}(S_1) \bigcap f^{-1}(S_2)\). Hence, we get \(f^{-1}(S) \subseteq f^{-1}(S_1) \bigcap f^{-1}(S_2)\).
• Now we shall show that \(f^{-1}(S) \supseteq f^{-1}(S_1) \bigcap f^{-1}(S_2)\). Let \(x \in f^{-1}(S_1) \bigcap f^{-1}(S_2)\). This means \(x \in f^{-1}(S_1)\) and \(x \in f^{-1}(S_2)\). This gives us \(f(x) \in S_1\) and \(f(x) \in S_2\). Hence, we get \(f(x) \in S_1 \bigcap S_2\).

Hence \(\tau_x\) is a topology. \(\Box\)

Topology without Tears - 1.1 Definitions and Propositions

Definition:

Let \(X\) be a non-empty set. A set \( \tau\) is of subsets of \(X\) is said to be a topology on \( X \) if

1. \(X\) and the empty set, \( \emptyset \), belong to \( \tau \)
2. The union of any (finite (or) infinite) number of sets in \( \tau\) belongs to \( \tau\)
3. The intersection of any two sets in \( \tau\) belongs to \( \tau\)

The pair \( (X,\tau) \) is called a topological space.

Definition:

Let \(X\) be any non-empty set and let \(\tau\) be the collection of all subsets of \(X\). Then \(\tau\) is called the discrete topology on the set \(X\). The topological space \( (X,\tau) \) is called a discrete space.

Definition:

Let \(X\) be any non-empty set and let \(\tau\) be \( \{X, \emptyset\} \). Then \(\tau\) is called the indiscrete topology on the set \(X\). The topological space \( (X,\tau) \) is called a indiscrete space.

Proposition:

If \( (X,\tau) \) is a topological space such that, for every \( x \in X\), the singleton set \( \{x\}\) is in \( \tau\), then \( \tau\) is the discrete topology.

Proof:

Consider any \(A \subseteq X\). Note that \( A = \{x \in X : x \in A \} = \bigcup_{x \in A} \{x\} \). Since \( \tau\) is a topology and by the definition of a topology, any union must belong to the topology and hence \( A = \bigcup_{x \in A} \{x\} \in \tau\). Hence, we get that any subset of \(X\) belongs to \(\tau\). Hence, \( \tau\) is a discrete topology. \(\Box\)

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
• \(\tau = \{\emptyset,\{1\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,2\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,3\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,4\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{1\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,2\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{2,3\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{2,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{2\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{2,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{4,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{3\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{2,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{3,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{4\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{1,2\},\{3,4\},X \} \)
• \(\tau = \{\emptyset,\{1,2\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{1,2\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{1,3\},\{2,4\},X \} \)
• \(\tau = \{\emptyset,\{1,3\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{1,3\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{1,4\},\{2,3\},X \} \)
• \(\tau = \{\emptyset,\{1,4\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{1,4\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2,3\},\{1,2,3\},X \} \)
• \(\tau = \{\emptyset,\{2,3\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{2,4\},\{1,2,4\},X \} \)
• \(\tau = \{\emptyset,\{2,4\},\{2,3,4\},X \} \)
• \(\tau = \{\emptyset,\{3,4\},\{1,3,4\},X \} \)
• \(\tau = \{\emptyset,\{3,4\},\{2,3,4\},X \} \)

Topology without Tears - 1.2 Definitions and Propositions

Definition:

Let \((X,\tau)\) be any topological space. Then the members of \(\tau\) are said to be open sets.

Proposition:

If \(X,\tau\) is any topological space, then

1. \(X\) and \(\emptyset\) are open sets.
2. The union of any (finite (or) infinite) number of open sets is an open set.
3. The intersection of any finite number of open sets is an open set.

Proof:

Clearly, \(1,2\) follow from the definitions of a topology. \(3\) follows from the definition of topology and problem \(4\) in the previous exercise. \(\Box\)

Definition:

Let \((X,\tau)\) be a topological space. A subset \(S \subset X\) is said to be a closed set in \((X,\tau)\) if its complement in \(X\), namely \(X \backslash S\), is open in \((X,\tau)\).

Proposition:

If \(X,\tau\) is any topological space, then

1. \(\emptyset\) and \(X\) are closed sets.
2. The intersection of any (finite (or) infinite) number of closed sets is a closed set.
3. The union of any finite number of closed sets is a closed set.

Proof:

• \(1\) follows immediately since \(X \backslash X = \emptyset \in \tau\) and \(X \backslash \emptyset = X \in \tau\).
• To prove \(2\), consider \(\bigcap_{\alpha \in \Gamma} A_{\alpha}\) where \(A_{\alpha}\) is a closed set i.e. \(X \backslash A_{\alpha} = X \bigcap A^c_{\alpha} \in \tau\). All we need to prove is \(X \bigcap \left(\bigcap_{\alpha \in \Gamma} A_{\alpha} \right)^c \in \tau\) i.e. we need to prove \(X \bigcap \left(\bigcup_{\alpha \in \Gamma}A_{\alpha}^c \right)\). Now since \(\tau\) is a topology and \(A_{\alpha}^c\) are open sets in \(\tau\), from property \(2\) of topology, we get \(\bigcup_{\alpha \in \Gamma} A_{\alpha}^c \in \tau\). Now from property \(3\) of topology, we get \(X \bigcap \left( \bigcup_{\alpha \in \Gamma} A_{\alpha}^c \right) \in \tau\). Hence, we get \(X \bigcap \left( \bigcap_{\alpha \in \Gamma} A_{\alpha}\right)^c \in \tau\) and hence, \(\bigcap_{\alpha \in \Gamma} A_{\alpha}\) is a closed set. Hence, the intersection of any (finite (or) infinite) number of closed sets is a closed set.
• To prove \(3\), consider closed sets \(\{A_k\}_{k=1}^{n}\). We need to prove that \(\bigcup_{k=1}^{n} A_k\) is also closed i.e. we need to prove that \(X \bigcap \left(\bigcup_{k=1}^{n} A_k \right)^c \in \tau\) i.e. we need to prove that \(X \bigcap \left( \bigcap_{k=1}^{n} A_k^c \right) \in \tau\) i.e. we need to prove that \(\bigcap_{k=1}^{n} \left( X \bigcap A_k^c \right) \in \tau\). Since \(A_k\) are closed by definition, we get \(X \bigcap A_{k}^c \in \tau\), \(\forall k \in \{1,2,\ldots,n\}\). Since \(\tau\) is a topology, using property \(3\) and problem \(4\) of the previous exercise, we get \(\bigcap_{k=1}^{n} \left( X \bigcap A_k^c \right) \in \tau\). Hence, the union of any finite number of closed sets is a closed set. \(\Box\)

Definition:

A subset \(S\) of a topological space \((X,\tau)\) is said to be clopen if it is both open and closed in \((X,\tau)\)

Topology without Tears - 1.1 Exercises

Exercises:

1. Let \( X = \{a,b,c,d,e,f \}\). Determine whether (or) not each of the following collection of subsets of \(X\) is a topology on \(X\)
• \( \tau_1 = \{X, \emptyset, \{a\}, \{a,f\}, \{b,f\}, \{a,b,f\} \} \)
• Note that \( \{a,f\} \bigcap \{b,f\} = \{f\} \notin \tau_1 \). Hence, \( \tau_1\) is not a topology.
• \( \tau_2 = \{X, \emptyset, \{a,b,f\}, \{a,b,d\}, \{a,b,d,f\} \} \)
• Again, note that \(\{a,b,f\}\bigcap \{a,b,d\} =\{a,b\} \notin \tau_2\). Hence, \( \tau_2\) is not a topology.
• \( \tau_3 = \{X, \emptyset, \{f\}, \{e,f\}, \{a,f\} \}\)
• Note that \(\{e,f \} \bigcup \{a,f\} = \{a,e,f\} \notin \tau_3\}\). Hence, \( \tau_3\) is not a topology.
2. Let \(X = \{a,b,c,d,e,f\}\). Which of the following collections of subsets of \(X\) is a topology on \(X\)? (Justify your answers)
• \(\tau_1 = \{X, \emptyset, \{c\}, \{b,d,e\}, \{b,c,d,e\}, \{b\} \}\)
• Not a topology since \( \{c\} \bigcup \{b\} = \{b,c\} \notin \tau_1\)
• \(\tau_2 = \{X, \emptyset, \{a\}, \{b,d,e\}, \{a,b,d\}, \{a,b,d,e\} \}\)
• Not a topology since \(\{b,d,e\}\bigcup \{a,b,d\} = \{b,d\} \notin \tau_2\)
• \(\tau_3 = \{X, \emptyset, \{b\}, \{a,b,c\}, \{d,e,f\}, \{b,d,e,f\} \}\)
• It is a topology since 
• It contains \(X\) and \(\emptyset\).
• Any union of sets in \(\tau_3\) is in \(\tau_3\)
• Any intersection of sets in \(\tau_3\) is in \(\tau_3\)
3. Let \(X = \{a,b,c,d,e,f \}\) and \(\tau\) is the discrete topology on \(X\). Which of the following statements are true?
• \(X \in \tau\) True
• \(\{X\} \in \tau\) False
• \(\{\emptyset\} \in \tau\) False
• \(\emptyset \in \tau\) True
• \(\emptyset \in X\) False
• \(\{\emptyset\} \in X\) False
• \(\{a\} \in \tau\) True
• \(a \in \tau\) False
• \(\emptyset \subseteq X\) True
• \(\{a\} \in X\) False
• \(\{\emptyset\} \subseteq X\) False
• \(a \in X\) True
• \(X \subseteq \tau\) False
• \(\{a\} \subseteq \tau\) False
• \(\{X\} \subseteq \tau\) True
• \(a \subseteq \tau\) False
4. Let \((X,\tau)\) be any topological space. Verify that the intersection of any finite number of members of \(\tau\) is a member of \(\tau\).
• Let \(P_n\) be the statement that "If \(\{A_i\}_{i=1}^n \in \tau \implies \bigcap_{i=1}^n A_i \in \tau \)". Let \(S = \{n \in \mathbb{N}: P_n \text{ is a true statement}\}\). From the definition of a topology, we have \(1 \in \mathbb{N}\). Assume that \(k \in S\) i.e. \(P_k\) is a true statement i.e. whenever \(\{A_i\}_{i=1}^k \in \tau \implies \bigcap_{i=1}^k A_i \in \tau \). Consider \(\{A_i\}_{i=1}^{k+1} \in \tau \). \( \bigcap_{i=1}^{k+1} A_i = \left( \bigcap_{i=1}^{k} A_i \right) \bigcap A_{k+1}\). By assumption, we have \(B = \left( \bigcap_{i=1}^{k} A_i \right) \in \tau\). Now from the definition of a topology, we get \(B \bigcap A_{k+1} \in \tau\) since \(B,A_{k+1} \in \tau\). Hence, we get \( \bigcap_{i=1}^{k+1} A_i \in \tau \) whenever \(\{A_i\}_{i=1}^{k+1} \in \tau \). Hence, \(P_{k+1}\) is a true statement. Hence, \( k+1 \in S\). So by principle of mathematical induction we have \( 1 \in S \), and \(k+1 \in S\) whenever \( k \in S \). Hence, \( S = \mathbb{N} \). Hence, the intersection of any finite number of members of \(\tau\) is a member of \(\tau\).
5. Let \(\mathbb{R}\) be the set of all real numbers. Prove that each of the following collections of subsets of \(\mathbb{R}\) is a topology
• \(\tau_1\) consists of \(\mathbb{R}, \emptyset\) and every interval \((-n,n)\), for any positive integer
• Clearly, \(\mathbb{R}, \emptyset \in \tau_1\)
• First note that \(\tau_1\) is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let \(A_{k} = (-k,k)\). Note that \(A_k\)'s are monotone increasing sequence of sets. Hence, any finite union of the form \(\bigcup_{l=1}^{m} A_{k_l} = A_{p}\) where \(p=\max(k_1,k_2,\ldots,k_m)\). And \(A_{p} \in \tau_1\) and hence any finite union again belongs to \(\tau_1\). Any infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_l \in \mathbb{N}\) can be rewritten either as a finite union of the form \(\bigcup_{l=1}^{m} A_{k_l}\) (or) an infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_i \neq k_j\). The claim now is that any infinite union where \(k_i \neq k_j\) is \(\mathbb{R}\) i.e. \(\bigcup_{l=1}^{\infty} A_{k_l} = \mathbb{R}\) where \(k_i \neq k_j\). To prove this, we prove the two way inclusion. Note that \(A_{k_l} \in \mathbb{R}\). Hence, \(\bigcup_{l=1}^{\infty} A_{k_l} \subseteq \mathbb{R}\). Further given any \(x \in \mathbb{R}\), by archimedian property \( \exists n \in \mathbb{N}\) such that \(x \in A_n\). Further since it is an infinite union, \( \exists k_l\) such that \( x \in A_n \subseteq A_{k_l}\). Hence, \(\bigcup_{l=1}^{\infty} A_{k_l} \supseteq \mathbb{R}\). Further, \(A_n \bigcup \mathbb{R} = \mathbb{R} \in \tau_1\) and \(A_n \bigcup \emptyset = A_n \in \tau_1\). Hence, any union is also in \(\tau_1\).
• Now we need to prove the last claim that intersection of any two elements of \(\tau_1\) gives an element in \(\tau_1\). Consider \(A_m,A_n \in \tau_1\). By well-ordering principle, we get \(m \lt n\) (or) \(m = n\) (or) \(m \gt n\). If \(m=n\), then \(A_m \bigcap A_n = A_n \in \tau_1\). If \(m \lt n\), then \(A_m \bigcap A_n = A_m \in \tau_1\). If \(m \gt n\), then \(A_m \bigcap A_n = A_n \in \tau_1\). Further, \(A_n \bigcap \mathbb{R} = A_n \in \tau_1\) and \(A_n \bigcap \emptyset = \emptyset \in \tau_1\). Hence, intersection of any two elements of \(\tau_1\) gives an element in \(\tau_1\).
• \(\tau_2\) consists of \(\mathbb{R}, \emptyset\) and every interval \([-n,n]\), for any positive integer
• Clearly, \(\mathbb{R}, \emptyset \in \tau_2\)
• First note that \(\tau_2\) is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let \(A_{k} = [-k,k]\). Note that \(A_k\)'s are monotone increasing sequence of sets. Hence, any finite union of the form \(\bigcup_{l=1}^{m} A_{k_l} = A_{p}\) where \(p=\max(k_1,k_2,\ldots,k_m)\). And \(A_{p} \in \tau_2\) and hence any finite union again belongs to \(\tau_2\). Any infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_l \in \mathbb{N}\) can be rewritten either as a finite union of the form \(\bigcup_{l=1}^{m} A_{k_l}\) (or) an infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_i \neq k_j\). The claim now is that any infinite union where \(k_i \neq k_j\) is \(\mathbb{R}\) i.e. \(\bigcup_{l=1}^{\infty} A_{k_l} = \mathbb{R}\) where \(k_i \neq k_j\). To prove this, we prove the two way inclusion. Note that \(A_{k_l} \in \mathbb{R}\). Hence, \(\bigcup_{l=1}^{\infty} A_{k_l} \subseteq \mathbb{R}\). Further given any \(x \in \mathbb{R}\), by archimedian property \( \exists n \in \mathbb{N}\) such that \(x \in A_n\). Further since it is an infinite union, \( \exists k_l\) such that \( x \in A_n \subseteq A_{k_l}\). Hence, \(\bigcup_{l=1}^{\infty} A_{k_l} \supseteq \mathbb{R}\). Further, \(A_n \bigcup \mathbb{R} = \mathbb{R} \in \tau_2\) and \(A_n \bigcup \emptyset = A_n \in \tau_2\). Hence, any union is also in \(\tau_2\).
• Now we need to prove the last claim that intersection of any two elements of \(\tau_2\) gives an element in \(\tau_2\). Consider \(A_m,A_n \in \tau_2\). By well-ordering principle, we get \(m \lt n\) (or) \(m = n\) (or) \(m \gt n\). If \(m=n\), then \(A_m \bigcap A_n = A_n \in \tau_2\). If \(m \lt n\), then \(A_m \bigcap A_n = A_m \in \tau_2\). If \(m \gt n\), then \(A_m \bigcap A_n = A_n \in \tau_2\). Further, \(A_n \bigcap \mathbb{R} = A_n \in \tau_2\) and \(A_n \bigcap \emptyset = \emptyset \in \tau_2\). Hence, intersection of any two elements of \(\tau_2\) gives an element in \(\tau_2\).
• \(\tau_3\) consists of \(\mathbb{R}, \emptyset\) and every interval \([n,\infty)\), for any positive integer
• Clearly, \(\mathbb{R}, \emptyset \in \tau_3\)
• First note that \(\tau_3\) is a countable set and hence all we are interested in is in finite (or) countably infinite unions. Let \(A_{k} = [k,\infty)\). Note that \(A_k\)'s are monotone decreasing sequence of sets. Hence, any finite union of the form \(\bigcup_{l=1}^{m} A_{k_l} = A_{p}\) where \(p=\min(k_1,k_2,\ldots,k_m)\). And \(A_{p} \in \tau_3\) and hence any finite union again belongs to \(\tau_3\). Any infinite union \(\bigcup_{l=1}^{\infty} A_{k_l}\) where \(k_l \in \mathbb{N}\) equals \(A_p\) where \(p = \min(k_1,k_2,\ldots) \). Further, \(A_n \bigcup \mathbb{R} = \mathbb{R} \in \tau_3\) and \(A_n \bigcup \emptyset = A_n \in \tau_3\). Hence, any union is also in \(\tau_3\).
• Now we need to prove the last claim that intersection of any two elements of \(\tau_3\) gives an element in \(\tau_3\). Consider \(A_m,A_n \in \tau_3\). By well-ordering principle, we get \(m \lt n\) (or) \(m = n\) (or) \(m \gt n\). If \(m=n\), then \(A_m \bigcap A_n = A_n \in \tau_3\). If \(m \lt n\), then \(A_m \bigcap A_n = A_n \in \tau_3\). If \(m \gt n\), then \(A_m \bigcap A_n = A_m \in \tau_3\). Further, \(A_n \bigcap \mathbb{R} = A_n \in \tau_3\) and \(A_n \bigcap \emptyset = \emptyset \in \tau_3\). Hence, intersection of any two elements of \(\tau_3\) gives an element in \(\tau_3\).
6. Let \(\mathbb{N}\) be the set of all positive integers. Prove that each of the following collections of subsets of \(\mathbb{N}\) is a topology.
• \(\tau_1\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{1,2,\ldots,n\}\) for any positive integer \(n\). This is called the initial segment topology.
• Argument similar to the above question
• \(\tau_2\) consists of \(\mathbb{N}, \emptyset\) and every set of the form \(\{n,n+1,\ldots\}\) for any positive integer \(n\). This is called the final segment topology.
• Argument similar to the above question
7. List all possible topologies on the following sets:
• \(X = \{a,b\}\)
• \(\tau_1 = \{\emptyset,X\} \)
• \(\tau_2 = \{\emptyset,\{a\},X\} \)
• \(\tau_3 = \{\emptyset,\{b\},X\} \)
• \(\tau_4 = \{\emptyset,\{a\},\{b\},X\} \)
• \(Y = \{a,b,c\}\)
• \(\tau_1 = \{\emptyset,X\} \)
• \(\tau_2 = \{\emptyset,\{a\},X\} \)
• \(\tau_3 = \{\emptyset,\{b\},X\} \)
• \(\tau_4 = \{\emptyset,\{c\},X\} \)
• \(\tau_5 = \{\emptyset,\{a\},\{b\},\{a,b\},X\} \)
• \(\tau_6 = \{\emptyset,\{c\},\{b\},\{c,b\},X\} \)
• \(\tau_7 = \{\emptyset,\{a\},\{c\},\{a,c\},X\} \)
• \(\tau_8 = \{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{b,c\},\{c,a\},X\} \)
8. Let \(X\) be an infinite set and \(\tau\) be a topology on \(X\). If every infinite subset of \(X\) is in \(\tau\), prove that \(\tau\) is the discrete topology.
• The idea is to prove that every singleton set is in the topology. By a proposition proved earlier, it follows that \(\tau\) is the discrete topology. To prove this, first observe that given an infinite set, \(X\), we can choose an infinite set \(A\) such that \(X \backslash A\) is also infinite.
• For instance, if the given set \(X\) is countably infinite, then we can set a bijection from \(\mathbb{N}\) to the set \(X\) i.e. we can write the set \(X\) as \(\{x_k: k \in \mathbb{N}\}\). Let \(A = \{x_k: k \in \mathbb{Z}^+_{odd}\}\). Clearly, both \(A\) and \(X \backslash A\) are countably infinite subsets of \(X\).
• If the given set \(X\) is uncountably infinite, then by axiom of choice, it is possible to choose a countably infinite subset, say \(A\) of \(X\). Hence, \(X \backslash A\) is also an infinite set, infact it is uncountably infinite. (I was initially unsure if this could be done. But thanks to math.stackexchange, Arturo Magidin told me that this was possible with the "Axiom of choice".)
• Now we have \(A\) and \(X \backslash A\) to be infinite sets. For any \(x \in X\), consider \(A \cup \{x\}\) and \(X \backslash A \cup \{x\}\). These two are infinite sets and hence both the sets are in the topology \(\tau\) and hence the intersection of the two sets must also be in the topology \(\tau\). The intersection of the above two sets is nothing but \(\{x\}\). This is true for every set \(\{x\}\). Hence, every singleton set belongs to the topology and hence the topology is discrete.
9. Let \(\mathbb{R}\) be the set of all real numbers. Precisely three of the following collections of subsets of \(\mathbb{R}\) are topologies? Identify these and justify your answer
• \(\tau\) consists of \(\mathbb{R}, \emptyset\) and every interval \((a,b)\), for \(a\) and \(b\) any real numbers with \(a \lt b\)
  • Not a topology since for instance \((1,2), (3,4) \in \tau\) but \((1,2) \bigcup (3,4) \notin \tau\)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r)\) for \(r \in \mathbb{R}\)
• Is a topology by argument similar to question \(6\)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r)\) for \(r \in \mathbb{Q}^+\)
• Not a topology since if we take rationals \(\{x_n\}_{n=1}^{\infty} \) such that \(x_{n+1} = \frac{2x_n}{2x_n^2+1} \) with \(x_1 = \frac12\) and let \(A_n = (-x_n,x_n)\) and consider \( \bigcup_{k=1}^{\infty}A_k \). Note that \(\{x_k\}_{k=1}^{\infty}\) is an increasing sequence bounded above and converges to \(\frac1{\sqrt{2}}\). This means that the sets \(A_k\) are monotone increasing sequence of sets, and \(A_k \in \tau\). \( \bigcup_{k=1}^{\infty}A_k = \left(-\frac1{\sqrt{2}},\frac1{\sqrt{2}} \right) \notin \tau \)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r]\) for \(r \in \mathbb{Q}^+\)
• Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1-\frac1n\). Let \(A_n = [-x_n,x_n]\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r)\) for \(r \in \mathbb{R}^+ \backslash \mathbb{Q} \)
• Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1- \frac{\sqrt{2}}{2n}\). Clearly, \(x_n \in \mathbb{R}^+ \backslash \mathbb{Q}\), \(\forall n \in \mathbb{N}\). Let \(A_n = (-x_n,x_n)\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r]\) for \(r \in \mathbb{R}^+ \backslash \mathbb{Q} \)
• Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = \sqrt{2}- \frac{1}{n}\). Clearly, \(x_n \in \mathbb{R}^+ \backslash \mathbb{Q}\), \(\forall n \in \mathbb{N}\). Let \(A_n = [-x_n,x_n]\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-\sqrt{2},\sqrt{2} \right) \notin \tau \)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r)\) for \(r \in \mathbb{R}^+ \)
• Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1- \frac1{n}\). Clearly, \(x_n \in \mathbb{R}^+\), \(\forall n \in \mathbb{N}\). Let \(A_n = [-x_n,x_n)\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \((-r,r]\) for \(r \in \mathbb{R}^+ \)
• Not a topology since if we consider \(\{x_n\}_{n=1}^{\infty}\) where \(x_n = 1- \frac1{n}\). Clearly, \(x_n \in \mathbb{R}^+\), \(\forall n \in \mathbb{N}\). Let \(A_n = (-x_n,x_n]\). Clearly, \(A_k \in \tau\) and \(A_k\) form a monotone increasing sequence of sets. \( \bigcup_{k=1}^{\infty}A_k = \left(-1,1 \right) \notin \tau \)
• \(\tau\) consists of \(\mathbb{R},\emptyset\) and every interval \([-r,r]\) and every interval \((-r,r)\) for \(r \in \mathbb{R}^+ \)
• Is a topology
• Clearly, \(X,\emptyset \in \tau\)
• Let \(A^{(0)}_{r}=(-r,r)\) and \(A^{(1)}_r = [-r,r]\). Consider \(A = \left (\bigcup_{x \in \Gamma_0} A^{(0)}_x \right) \bigcup \left( \bigcup_{y \in \Gamma_1} A^{(1)}_y \right) \). Then \(A = A^{(0)}_x\) for some \(x \in \mathbb{R}^+\) (or) \(A = A^{(1)}_y\) for some \(y \in \mathbb{R}^+\) (or) \(A=\mathbb{R}\).
• Intersection of any two sets is closed in \(\tau\)
• \(\tau\) consists of \(\mathbb{R}, \emptyset\), every interval \([-n,n]\), and every interval \((-r,r)\), for \(n \in \mathbb{Z}^+\) and \(r \in \mathbb{R}^+\)
• Is a topology. Again check the definitions.