Continuous Functions

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Continuous Functions

We now define the notion of a continuous function. Intuitively, this can be thought of as a map that preserves topological structure, just as a homomorphism does in an algebraic setting.

Definition 1 A function \(f:X\rightarrow Y\) between topological spaces \(X\), \(Y\) is said to be continuous at \(x\in X\) if for every neighborhood \(V\) of \(f(x)\in Y\), there exists a neighborhood \(U\) of \(x\) such that \(f(U)\subseteq V\).

For any function \(f:X\rightarrow Y\) between sets \(X,Y\) and any \(U\subseteq X\), we have \(U\subseteq f^{-1}(f(U))\). Thus, if any subset \(V\subseteq Y\) satisfies \(f(U)\subseteq V\), then

\[U\subseteq f^{-1}(f(U))\subseteq f^{-1}(V)\]

holds. Therefore, to prove that a function \(f:X\rightarrow Y\) between two topological spaces is continuous at \(x\in X\), it suffices to show that for every neighborhood \(V\) of \(f(x)\in Y\), the set \(f^{-1}(V)\) is a neighborhood of \(x\); more generally, it suffices to show that for a fixed local base \(\mathcal{B}(f(x))\) of \(f(x)\) and any \(V\in\mathcal{B}(f(x))\), we have \(f^{-1}(V)\in\mathcal{N}(x)\).

Proposition 2 Let \(f:X\rightarrow Y\) be a function between two topological spaces that is continuous at a point \(x\). If \(x\in\cl(A)\) for some \(A\subseteq X\), then \(f(x)\in\cl(f(A))\).

Proof

Choose any neighborhood \(V\) of \(f(x)\in Y\). Then \(f^{-1}(V)\) is a neighborhood of \(x\), so \(f^{-1}(V)\cap A\neq\emptyset\) (§Interior, Closure, and Boundary, ⁋Proposition 6). Taking \(x'\in f^{-1}(V)\cap A\), we have \(f(x')\in V\cap f(A)\). In particular, \(V\cap f(A)\neq\emptyset\), so applying §Interior, Closure, and Boundary, ⁋Proposition 6 again yields \(f(x)\in\cl(f(A))\).

Proposition 3 Let topological spaces \(X,Y,Z\) be given. If \(f:X\rightarrow Y\) is continuous at a point \(x\in X\) and \(g:Y\rightarrow Z\) is continuous at \(f(x)\), then the composition \(g\circ f\) is also continuous.

Proof

Choose any neighborhood \(W\) of \((g\circ f)(x)\). Since \(g\) is continuous at \(f(x)\), the set \(g^{-1}(W)\) is a neighborhood of \(f(x)\). Since \(f\) is continuous at \(x\), the set \(f^{-1}(g^{-1}(W))\) is a neighborhood of \(x\). (§Operations on Binary Relations, ⁋Proposition 6)

If \(f\) is continuous at every point of \(X\), we call \(f\) a continuous function. The following theorem presents several equivalent characterizations of continuity.

Theorem 4 For two topological spaces \(X,Y\) and a function \(f:X\rightarrow Y\), the following are all equivalent.

1. \(f\) is continuous.
2. For any subset \(A\) of \(X\), we have \(f(\cl A)\subset\cl f(A)\).
3. For any closed set \(C\) in \(Y\), the set \(f^{-1}(C)\) is closed in \(X\).
4. For any open set \(V\) in \(Y\), the set \(f^{-1}(V)\) is open in \(X\).

Proof

That the first condition implies the second is a consequence of Proposition 2.

Now assume the second condition and prove the third. For any closed set \(C\) in \(Y\), the inclusion

\[f(\cl(f^{-1}(C))\subseteq \cl(f(f^{-1}(C))\subseteq\cl(C)=C\]

holds, so from

\[\cl(f^{-1}(C))\subseteq f^{-1}(f(\cl(f^{-1}(C)))\subseteq f^{-1}(C)\]

we see that \(f^{-1}(C)\) is closed. Since the identity \((f^{-1}(A))^c=f^{-1}(A^c)\) holds for any subset \(A\subseteq Y\), it is obvious that the fourth condition follows from this.

Thus it suffices to assume the fourth condition and prove the first. Choose any \(x\in X\), and let \(V\) be any neighborhood of \(f(x)\in Y\). Then there exists an open neighborhood \(V'\) of \(f(x)\) with \(f(x)\in V'\subseteq V\). By the fourth condition, \(f^{-1}(V')\) is an open neighborhood of \(x\in X\), and from \(f^{-1}(V')\subseteq f(V)\) we see that \(f(V)\) is a neighborhood of \(x\).

By Proposition 3, if two continuous functions \(f:X\rightarrow Y\) and \(g:Y\rightarrow Z\) are given, then \(g\circ f\) is also continuous.

Let a continuous function \(f:X\rightarrow Y\) between two topological spaces \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be given. Then for any \(V\in\mathcal{T}_Y\), we have \(f^{-1}(V)\in\mathcal{T}_X\), so the formula

\[f^\mathcal{T}(V):=f^{-1}(V),\qquad V\in\mathcal{T}_Y\]

well-defines a function \(f^\mathcal{T}:\mathcal{T}_Y\rightarrow\mathcal{T}_X\).

Now assume that \(f\) is a bijection and consider two distinct elements \(V_1,V_2\) of \(\mathcal{T}_Y\). Without loss of generality, suppose \(y\in V_1\setminus V_2\); then

\[f^{-1}(y)\in f^\mathcal{T}(V_1)\setminus f^\mathcal{T}(V_2)\]

holds, so \(f^{\mathcal{T}}\) is injective.

Example 5 In general, \(f^{\mathcal{T}}\) need not be surjective. For instance, let \(X_1\) be the space \(\mathbb{N}\) with the discrete topology \(\mathcal{T}_1\), and let \(X_2\) be the space \(\mathbb{N}\) with the trivial topology \(\mathcal{T}_2\). Considering the identity map \(\id:\mathbb{N}\rightarrow\mathbb{N}\) as a set map, \(\id\) is a continuous bijection, but the function

\[\id^\mathcal{T}:\mathcal{T}_2\rightarrow\mathcal{T}_1\]

cannot be surjective. (§Operations on Cardinals, ⁋Proposition 10 (Cantor))

However, if the inverse function \(f^{-1}\) of a bijection \(f\) is also continuous, then \((f^{-1})^\mathcal{T}:\mathcal{T}_X\rightarrow\mathcal{T}_Y\) is well-defined, and from the definition it is obvious that

\[f^\mathcal{T}\circ (f^{-1})^\mathcal{T}=\id_{\mathcal{T}_X},\qquad (f^{-1})^\mathcal{T}\circ f^\mathcal{T}=\id_{\mathcal{T}_Y}\]

so \(f^\mathcal{T}\) is also bijective. We define such a situation as follows.

Definition 6 A continuous function \(f:X\rightarrow Y\) is called a homeomorphism if there exists another continuous function \(g:Y\rightarrow X\) such that \(f\circ g=\id_Y\) and \(g\circ f=\id_X\).

That is, two topological spaces \(X,Y\) are homeomorphic means not only that there exists a bijection between them as sets, but also that this bijection acts on the open sets of the two sets in exactly the same way. An example of a continuous bijection that is not a homeomorphism is precisely Example 5 above.

---

References

[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.

---