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. \(\Box\)