콘텐츠로 이동

§ 18. Continuous Functions

Definition 18.1 : Continuous Function

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

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 wish to emphasize this fact, we can say that "\(f\) is continuous relative to specific topologies on \(X\) and \(Y\)".

Theorem 18.0

If the topology of the range space \(Y\) is given by a basis \(\mathcal{B}\), then to prove continuity of \(f\) it suffices to show that the inverse image of every basis element 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.

Proof

If the topology of the range space \(Y\) is given by a basis \(\mathcal{B}\), the arbitrary open set \(V\) of \(Y\) can be written as a union of basis elements

\[ V=\bigcup_{\alpha \in J} B_{\alpha} \]

Then

\[ f^{-1}(V)=\bigcup_{\alpha \in J} f^{-1}\left(B_{\alpha}\right) \]

so that \(f^{-1}(V)\) is open if each set \(f^{-1}\left(B_{\alpha}\right)\) is open.

If the topology on \(Y\) is given by a subbasis \(\mathcal{S}\), 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

\[ f^{-1}(B)=f^{-1}\left(S_{1}\right) \cap \cdots \cap f^{-1}\left(S_{n}\right) \]

that the inverse image of every basis element is open.

Example

Let us consider a function like those studied in analysis, a "real-valued function of a real variable",

\[ f: \mathbb{R} \longrightarrow \mathbb{R} \]

In analysis, one defines continuity of \(f\) via the " \(\epsilon-\delta\) definition". As one would expect, the \(\epsilon-\delta\) definition and ours are equivalent. To prove that our definition implies the \(\epsilon-\delta\) definition, for instance, we proceed as follows:

Given \(x_{0}\) in \(\mathbb{R}\), and given \(\epsilon>0\), the interval \(V=\left(f\left(x_{0}\right)-\epsilon, f\left(x_{0}\right)+\epsilon\right)\) is an open set of the range space \(\mathbb{R}\). Therefore, \(f^{-1}(V)\) is an open set in the domain space \(\mathbb{R}\). Because \(f^{-1}(V)\) contains the point \(x_{0}\), it contains some basis element \((a, b)\) about \(x_{0}\). We choose \(\delta\) to be the smaller of the two numbers \(x_{0}-a\) and \(b-x_{0}\). Then if \(\left|x-x_{0}\right|<\delta\), the point \(x\) must be in \((a, b)\), so that \(f(x) \in V\), and \(\left|f(x)-f\left(x_{0}\right)\right|<\epsilon\), as desired.

Proving that the \(\epsilon-\delta\) definition implies our definition is similar.

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 \(f(\bar{A}) \subset \overline{f(A)}\).

  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 \(f(x) \in \overline{f(A)}\). 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 \(f(x) \in \overline{f(A)}\), 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 \(f(A)=\) \(f\left(f^{-1}(B)\right) \subset B\). Therefore, if \(x \in \bar{A}\),

\[ f(x) \in f(\bar{A}) \subset \overline{f(A)} \subset \bar{B}=B, \]

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 \(f\left(U_{x}\right) \subset V\). 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.

Definition 18.2 : Homeomorphism / Topological Property

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.

Another way to define a homeomorphism is to 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 \(X\).

Homeomorphism is a bijective correspondence that preserves the topological structure involved.

Topological Imbedding

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 \(f^{\prime}: X \rightarrow Z\) obtained by restricting the range of \(f\) is bijective. If \(f^{\prime}\) happens to be a homeomorphism of \(X\) with \(Z\), we say that the map \(f: X \rightarrow Y\) is a topological imbedding, or simply an imbedding, of \(X\) in \(Y\).

Theorem 18.2 : Rules for Constructing Continuous Functions

Let \(X, Y\), and \(Z\) be topological spaces.

  1. (Constant function) If \(f: X \rightarrow Y\) maps all of \(X\) into the single point \(y_{0}\) of \(Y\), then \(f\) is continuous.

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

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

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

  5. (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.

  6. (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 \(f \mid U_{\alpha}\) is continuous for each \(\alpha\).

Proof

  1. 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.

  2. If \(U\) is open in \(X\), then \(j^{-1}(U)=U \cap A\), which is open in \(A\) by definition of the subspace topology.

  3. If \(U\) is open in \(Z\), then \(g^{-1}(U)\) is open in \(Y\) and \(f^{-1}\left(g^{-1}(U)\right)\) is open in \(X\). But

    \[ f^{-1}\left(g^{-1}(U)\right)=(g \circ f)^{-1}(U), \]

    by elementary set theory.

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

  5. 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 \(f^{-1}(U)\) 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 \(j: Y \rightarrow Z\).

  6. By hypothesis, we can write \(X\) as a union of open sets \(U_{\alpha}\), such that \(f \mid U_{\alpha}\), is continuous for each \(\alpha\). Let \(V\) be an open set in \(Y\). Then

    \[ f^{-1}(V) \cap U_{\alpha}=\left(f \mid U_{\alpha}\right)^{-1}(V), \]

    because both expressions represent the set of those points \(x\) lying in \(U_{\alpha}\) for which \(f(x) \in V\). Since \(f \mid U\) is continuous, this set is open in \(U_{\alpha}\), and hence open in \(X\). But

    \[ f^{-1}(V)=\bigcup_{\alpha}\left(f^{-1}(V) \cap U_{\alpha}\right), \]

    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 \(x \in B\).

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 Theorem 18.2.

Proof

Let \(C\) be a closed subset of \(Y\). Now

\[ h^{-1}(C)=f^{-1}(C) \cup g^{-1}(C) \]

by 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 \(X\).

Theorem 18.4 : Maps into Products

Let \(f: A \rightarrow X \times Y\) be given by the equation

\[ f(a)=\left(f_{1}(a), f_{2}(a)\right) . \]

Then \(f\) is continuous if and only if the functions

\[ f_{1}: A \longrightarrow X \quad \text { and } \quad f_{2}: A \longrightarrow Y \]

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 \(\pi_{2}^{-1}(V)=X \times V\), and these sets are open if \(U\) and \(V\) are open. Note that for each \(a \in A\),

\[ f_{1}(a)=\pi_{1}(f(a)) \quad \text { and } \quad f_{2}(a)=\pi_{2}(f(a)). \]

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 show 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.