[Munkres Topology] 18. Continuous Functions

 This article is one of the posts in the Textbook Commentary Project.

The concept of continuous function is basic to much of mathematics. In this section, we shall formulate a definition of continuity that will include all these as special cases, and we shall study various properties of continuous functions. Many of these properties are direct generalizations of things you learned about continuous functions in calculus and analysis.

Continuity of a Function

Let $X$ and $Y$ be topological spaces. A function $f: X\rightarrow Y$ is said to be continuous if for each open subset $V$ on $Y$, the set $f^{-1}(V)$ is an open subset of $X$.

Recall that $f^{-1}(V)$ is the set of all points $x$ of $X$ for which $f(x)\in V$; it is empty if $V$ does not intersect the image set $f(X)$ of $f$.

Continuity of a function depends not only upon the function $f$ itself, but also on the topologies specified for its domain and range. If we with to emphasize this fact, we can say that $f$ is continuous relative to specific topologies on $X$ and $Y$.

Let us note that if the topology of the range space $Y$ is given by a basis $\mathbb{B}$, then to prove continuity of $f$ it suffices to show that the inverse image of every basis element is open: The arbitrary open set $V$ of $Y$ can be written as a union of basis elements $$V=\bigcup_{\alpha\in J} B_\alpha.$$ Then $$f^{-1}(V)=\bigcup_{\alpha\in J} f^{-1} (B_\alpha),$$ so that $f^{-1}(V)$ is open if each set $f^{-1}(B_\alpha)$ is open.

If the topology on $Y$ is given by a subbasis $\mathcal{S}$, to prove continuity of $f$ it will even suffice to show that the inverse image of each subbasis element is open: The arbitrary basis element $B$ for $Y$ can be written as a finite intersection $S_1\cap \cdots \cap S_n$ of subbasis elements; it follows from the equation $$f^{-1}(B)=f^{-1}(S_1)\cap \cdots \cap f^{-1}(S_n)$$ that the inverse image of every basis element is open.

In analysis, one studies several different but equivalent ways of formulating the definition of continuity. Some of these generalize to arbitrary spaces, and they are considered in the theorems that follow. The familiar '$\epsilon - \delta$' definition and the 'convergent sequence definition' do not generalize to arbitrary spaces; they will be treated when we study metric spaces.

Theorem 18.1. Let $X$ and $Y$ be topological spaces; let $f:X\rightarrow Y$. Then the following are equivalent: 

(1) $f$ is continuous.

(2) For every subset $A$ of $X$, one has $f(\bar{A})\subset \bar{f(A)}$.

(3) For every closed set $B$ of $Y$, the set $f^{-1}(B)$ is closed in $X$.

(4) For each $x \in X$ and each neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ such that $f(U)\subset V$.

If the condition in (4) holds for the point $x$ of $X$, we say that $f$ is continuous at the point $x$.

Proof. We show that $(1)\Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$ and that $(1)\Rightarrow (4) \Rightarrow (1)$.

[$(1)\Rightarrow (2)$] Assume that $f$ is continuous. Let $A$ be a subset of $X$. We show that if $x\in \bar{A}$, then $f(x)\in \bar{f(A)}$. Let $V$ be a neighborhood of $f(x)$. Then $f^{-1}(V)$ is an open set of $X$ containing $x$; it must intersect $A$ in some point $y$. Then $V$ intersects $f(A)$ in the point $f(y)$, so that $f(x)\in \bar{f(A)}$, as desired.

[$(2)\Rightarrow (3)$] Let $B$ be closed in $Y$ and let $A=f^{-1}(B)$, We wish to prove that $A$ is closed in $X$; we show that $\bar{A}=A$. By elementary set theory, we have $f(A)=f(f^{-1}(B))\subset B$. Therefore, if $x\in \bar{A}$, $$f(x)\in f(\bar{A})\subset \bar{f(A)}\subset \bar{B}=B,$$ so that $x \in f^{-1}(B)=A$. Thus $\bar{A}\subset A$, so that $\bar{A}=A$, as desired.

[$(3)\Rightarrow (1)$] Let $V$ be an open set of $Y$. Set $B=Y-V$. Then $$f^{-1}(B)=f^{-1}(Y)-f^{-1}(V)=X-f^{-1}(V).$$ Now $B$ is a closed set of $Y$. Then $f^{-1}(B)$ is closed in $X$ by hypothesis, so that $f^{-1}(V)$ is open in $X$, as desired.

[$(1)\Rightarrow (4)$] Let $x\in X$ and let $V$ be a neighborhood of $f(x)$. Then the set $U=f^{-1}(V)$ is a neighborhood of $x$ such that $f(U)\subset V$.

[$(4)\Rightarrow (1)$] Let $V$ be an open set of $Y$; let $x$ be a point of $f^{-1}(V)$. Then $f(x)\in V$, so that by hypothesis there is a neighborhood $U_x$ of $x$ such that $f(U_x)\subset V$. Then $U_x \subset f^{-1}(V)$. It follows that $f^{-1}(V)$ can be written as the union of the open sets $U_x$, so that it is open.

Homeomorphisms

Let $X$ and $Y$ be topological spaces; let $f:X\rightarrow Y$ be a bijection. If both the function $f$ and the inverse function $$f^{-1}:Y\rightarrow X$$ are continuous, then $f$ is called a homeomorphism.

The condition that $f^{-1}$ be continuous says that for each open set $U$ of $X$, the inverse image of $U$ under the map $f^{-1}:Y\rightarrow X$ is open in $Y$. But the inverse image of $U$ under the map $f^{-1}$ is the same as the image of $U$ under the map $f$. See Figure 18.1. So another way to define a homeomorphism is say that it is a bijective correspondence $f:X\rightarrow Y$ such that $f(U)$ is open if and only if $U$ is open.

This remark shows that a homeomorphism $f:X\rightarrow Y$ gives us a bijective correspondence not only between $X$ and $Y$ but between the collections of open sets of $X$ and of $Y$. As a result, any property of $X$ that is entirely expressed in terms of the topology of $X$ (that is, in terms of the open sets of $X$) yields, via the correspondence $f$, the corresponding property for the space $Y$. Such a property of $X$ is called a topological property of $X$. 

Now suppose that $f:X\rightarrow Y$ is an injective continuous map, where $X$ and $Y$ are topological spaces. Let $Z$ be the image set $f(X)$, considered as a subspace of $Y$; then the function $f':X\rightarrow Z$ obtained by restricting the range of $f$ is bijective. If $f'$ happens to be a homoemorphism of $X$ with $Z$, we say that the map $f:X\rightarrow Y$ is a topological imbedding, or simply an imbedding, of $X$ in $Y$.

Constructing Continuous Functions

Theorem 18.2 (Rules for constructing continuous functions). Let $X$, $Y$, and $Z$ be topological spaces.

(a) (Constant function) If $f:X\rightarrow Y$ maps all of $X$ into the single point $y_0$ of $Y$, then $f$ is continuous.

(b) (Inclusion) If $A$ is a subspace of $X$, the inclusion function $j:A\rightarrow X$ is continuous.

(c) (Composites) If $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are continuous, then the map $g\circ f: X\rightarrow Z$ is continuous.

(d) (Restricting the domain) If $f:X\rightarrow Y$ is continuous, and if $A$ is a subspace of $X$, then the restricted function $f|A: A\rightarrow Y$ is continuous.

(c) (Restricting or expanding the range) Let $f:X\rightarrow Y$ be continuous. If $Z$ is a subspace of $Y$ containing the image set $f(X)$, then the function $g:X\rightarrow Z$ obtained by restricting the range of $f$ is continuous. If $Z$ is a space having $Y$ as a subspace, then the function $h:X\rightarrow Z$ obtained by expanding the range of $f$ is continuous.

(f) (Local formulation of continuity) The map $f:X\rightarrow Y$ is continuous if $X$ can be written as the union of open sets $U_\alpha$ such that $f|U_\alpha$ is continuous for each $\alpha$.

Proof.

(a) Let $f(x)=y_0$ for every $x$ in $X$. Let $V$ be open in $Y$. The set $f^{-1}(V)$ equals $X$ or $\varnothing$, depending on whether $V$ contains $y_0$ or not. In either case, it is open.

(b) If $U$ is open in $Z$, then $g^{-1}(U)$ is open in $Y$ and $f^{-1}(g^{-1}(U))$ is open in $X$. But $$f^{-1}(g^{-1}(U))=(g\circ f)^{-1}(U),$$ by elementary set theory.

(d) The function $f|A$ equals the composite of the inclusion map $j:A\rightarrow X$ and the map $f: X\rightarrow Y$, both of which are continuous.

(e) Let $f:X\rightarrow Y$ be continuous. If $f(X)\subset Z \subset Y$, we show that the function $g:X\rightarrow Z$ obtained from $f$ is continuous. Let $B$ be open in $Z$. Then $B=Z\cap U$ for some open set $U$ of $Y$. Because $Z$ contains the entire image set $f(X)$, $$f^{-1}(U)=g^{-1}(B),$$ by elementary set theory. Since $f^{-1}$ is open, so is $g^{-1}(B)$.

To show $h:X\rightarrow Z$ is continuous if $Z$ has $Y$ as a subspace, note that $h$ is the composite of the map $f:X\rightarrow Y$ and the inclusion map $j:Y\rightarrow Z$.

(f) By hypothesis, we can write $X$ as a union of open sets $U_\alpha$, such that $f|U_\alpha$, is continuous for each $\alpha$. Let $V$ be an open set in $Y$. Then $$f^{-1}(V)\cap U_\alpha=(f|U_\alpha)^{-1}(V),$$ because both expressions represent the set of those point $x$ lying in $U_\alpha$ for which $f(x)\in V$. Since $f|U$ is continuous, this set is open in $U_\alpha$, and hence open in $X$. But $$f^{-1}(V)=\bigcup_\alpha (f^{-1}(V)\cap U_\alpha),$$ so that $f^{-1}(V)$ is also open in $X$.

Theorem 18.3 (The pasting lemma). Let $X=A\cup B$, where $A$ and $B$ are closed in $X$. Let $f:A\rightarrow Y$ and $g:B\rightarrow Y$ be continuous. If $f(x)=g(x)$ for every $x\in A\cap B$, then $f$ and $g$ combine to give a continuous function $h:X\rightarrow Y$, defined by setting $h(x)=f(x)$ if $x\in A$, and $h(x)=g(x)$ if $x\in B$.

Proof. Let $C$ be a closed subset of $Y$. Now $$h^{-1}(C)=f^{-1}(C)\cup g^{-1}(C),$$ be elementary set theory. Since $f$ is continuous, $f^{-1}(C)$ is closed in $A$ and, therefore, closed in $X$. Similarly, $g^{-1}(C)$ is closed in $B$ and therefore closed in $X$. Their union $h^{-1}(C)$ is thus closed in $X$.

This theorem also holds if $A$ and $B$ are open sets in $X$; this is just a special case of the 'local formulation of continuity' rule given in preceding theorem.

Theorem 18.4 (Maps into products). Let $f:A\rightarrow X\times Y$ be given by the equation $$f(a)=(f_1(a),f_2(a)).$$ Then $f$ is continuous if and only if the functions $$f_1: A\rightarrow X\quad \mbox{and} \quad f_2:A\rightarrow Y$$ are continuous.

The maps $f_1$ and $f_2$ are called the coordinate functions of $f$.

Proof. Let $\pi_1: X\times Y\rightarrow X$ and $\pi_2:X\times Y\rightarrow Y$ be projections onto the first and second factors, respectively. These maps are continuous. For $\pi_1^{-1}(U)=U\times Y$ and $\pi_2^{-1}(V)=X\times V$m and these sets are open if $U$ and $V$ are open. Note that for each $a\in A$, $$f_1(a)=\pi_1(f(a))\quad \mbox{and} \quad f_2(a)=\pi_2(f(a)).$$

If the function $f$ is continuous, then $f_1$ and $f_2$ are composites of continuous functions and therefore continuous. Conversely, suppose that $f_1$ and $f_2$ are continuous. We shoe that for each basis element $U\times V$ for the topology of $X\times Y$, its inverse image $f^{-1}(U\times V)$ is open. A point $a$ is in $f^{-1}(U\times V)$ if and only if $f(a)\in U\times V$, that is, if and only if $f_1(a)\in U$ and $f_2(a)\in V$. Therefore, $$f^{-1}(U\times V)=f^{-1}_1(U)\cap f_2^{-1}(V).$$

Since both of the sets $f_1^{-1}(U)$ and $f_2^{-1}(V)$ are open, so is their intersection.

Yet another method for constructing continuous functions that is familiar from analysis is to take the limit of an infinite sequence of functions. There is a theorem to the effect that if a sequence of continuous real-valued functions of a real variable converges uniformly to a limit function, then the limit function is necessarily continuous. This theorem is called the Uniform Limit Theorem. It is used, for instance, to demonstrate the continuity of the trigonometric functions, when one defines these functions rigorously using the infinite series definitions of the sine and cosine. This theorem generalizes to a theorem about maps of an arbitrary topological space X into a metric space Y. We shall prove it in Chapter 21.