Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). We say that \(f\) is continuous if \(f^{-1}\bof U\) is open in \(X\) for every open \(U\sub Y\).
If \(X\) is discrete, then any function with domain \(X\) into any topological space \(Y\) is continuous.
If \(Y\) has the trivial topology then any function \(f:X\to Y\) for any topological space \(X\) is discrete.
If \(X\) and \(Y\) are any topological spaces and \(f:X\to Y\) is constant, then it is continuous.
If \(X\) is a topological space and \(f:X\to X\) is the identity function (\(f\of x=x\) for each \(x\in X\)), then \(f\) is continuous.
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). The following conditions are equivalent:
1. \(f\) is continuous.
2. \(f^{-1}\bof C\) is closed in \(X\) for any closed \(C\sub Y\).
3. \(f\bof {\ob A}\sub \ob {f\bof A}\) for any \(A\sub X\).
4. \(\ob {f^{-1}\bof B}\sub f^{-1}\bof {\ob B}\) for any \(B\sub Y\).
Proof. Assume 1. and let \(C\sub Y\) be closed. Then \(Y\sem C\) is open in \(Y\) so \(f^{-1}\bof {Y\sem C}\) is open in \(X\). Since
\[ f^{-1}\bof C=X\sem f^{-1}\bof {Y\sem C}, \]
it follows that \(f^{-1}\bof C\) is closed in \(X\). Thus 2. holds. Similarly, 2. implies 1.
Now we show that 1. implies 3. Let \(A\sub X\) and \(y\in f\bof {\ob A}\). Then \(y=f\of x\) for some \(x\in \ob A\). Let \(U\) be an open nbhd of \(y\) in \(Y\). Then \(f^{-1}\bof U\) is an open nbhd of \(x\) so
\[ f^{-1}\bof U\cap A\neq \emp . \]
Thus there is \(z\in A\) with \(f\of z\in U\) so
\[ f\bof A\cap U\neq \emp . \]
It follows that \(y\in \ob {f\bof A}\) and so 3. holds.
Now we show that 3. implies 4. Let \(B\sub Y\). With \(A:=f^{-1}\bof B\), 3. implies that
\[ f\bof {\ob {f^{-1}\bof B}}\sub \ob {f\bof {f^{-1}\bof B}}. \]
Since \(f\bof {f^{-1}\bof B}\sub B\), it follows that \(f\bof {\ob {f^{-1}\bof B}}\sub \ob B\), so
\[ \ob {f^{-1}\bof B}\sub f^{-1}\bof {\ob B}. \]
It remains to show that 4. implies 2. Let \(C\sub Y\) be closed. Then 4. implies that
\[ \ob {f^{-1}\bof C}\sub f^{-1}\bof {\ob C}=f^{-1}\bof C. \]
Since \(f^{-1}\bof C\sub \ob {f^{-1}\bof C}\), it follows that \(f^{-1}\bof C\) is closed. Thus 2. holds. □
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). Let \(\sb \) be a basis and \(\ss \) be a subbasis for the topology on \(Y\). The following conditions are equivalent:
1. \(f\) is continuous.
2. \(f^{-1}\bof B\) is open in \(X\) for every \(B\in \sb \).
3. \(f^{-1}\bof S\) is open in \(X\) for every \(S\in \ss \).
Proof. Since every member of \(\sb \) and every member of \(\ss \) is open in \(Y\), 1. implies both 2. and 3. If 2. holds, and \(U\) is open in \(Y\), then \(U=\bigcup \sa \) for some \(\sa \sub \sb \). Then
\[ f^{-1}\bof U=\bigcup _{B\in \sa }f^{-1}\bof B \]
is open in \(X\) so 1. holds.
Now assume 3. Let \(U\) be open in \(Y\). If \(U=Y\), then \(f^{-1}\bof U=X\) is open in \(X\). If \(U=\emp \), then \(f^{-1}\bof U=\emp \) is open in \(X\). Otherwise, \(U=\bigcup \sa \) for some family \(\sa \) consisting of intersections of finite nonempty subfamilies of \(\ss \). Since
\[ f^{-1}\bof U=\bigcup _{A\in \sa }f^{-1}\bof A, \]
it suffices to show that \(f^{-1}\bof A\) is open for every \(A\in \sa \). If
\[ A:=S_{1}\cap S_{2}\cap \ds \cap S_{k}, \]
then
\[ f^{-1}\bof A=f^{-1}\bof {S_{1}}\cap f^{-1}\bof {S_{2}}\cap \ds \cap f^{-1}\bof {S_{k}}, \]
which is an intersection of finitely many open sets in \(X\). Thus \(f^{-1}\bof A\) is open in \(X\). □
Let \(X,Y,Z\) be topological space and \(f:X\to Y\) and \(g:Y\to Z\) be continuous. Then \(g\circ f:X\to Y\) is continuous.
Proof. Let \(U\) be open in \(Z\). Then
\[ \cur {g\circ f}^{-1}\bof U=g^{-1}\bof {f^{-1}\bof U}. \]
Since \(f\) is continuous, \(f^{-1}\bof U\) is open in \(Y\) and since \(g\) is continuous, \(g^{-1}\bof {f^{-1}\bof U}\) is open in \(X\). Thus \(g\circ f\) is continuous. □
Let \(X\) be a topological space and \(Y\) be a subset of \(X\). The subspace topology on \(Y\) is the smallest topology on \(Y\) for which the embedding \(j:Y\to X\) (with \(j\of y:=y\) for each \(y\in Y\)) is continuous.
Proof. If \(U\) is open in \(X\), then \(j^{-1}\bof U=U\cap Y\) is open in the subspace topology on \(Y\). Thus \(j\) is continuous. Assume that \(\st \) is any topology on \(Y\) for which \(j\) is continuous. Then
\[ U\cap Y=j^{-1}\bof U\in \st \]
for any open \(U\sub X\) so \(\st \) is larger than the subspace topology on \(Y\). □
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). Then \(f\) is continuous at \(x\in X\) provided \(f^{-1}\bof U\) is a nbhd of \(x\) for any nbhd \(U\) of \(f\of x\).
Let \(X\) and \(Y\) be topological spaces. A function \(f:X\to Y\) is continuous if and only if \(f\) is continuous at each \(x\in X\).
Proof. Assume that \(f\) is continuous and \(x\in X\). If \(U\) is a nbhd of \(f\of x\), then there is open \(U’\) in \(Y\) with \(f\of x\in U’\sub U\). Since \(f\) is continuous, \(f^{-1}\bof {U’}\) is open in \(X\) and
\[ x\in f^{-1}\bof {U’}\sub f^{-1}\bof U. \]
Thus \(f^{-1}\bof U\) is a nbhd of \(x\) in \(X\).
Now assume that \(f\) is continuous at each \(x\in X\). Let \(U\) be open in \(Y\). For each \(x\in f^{-1}\bof U\), the set \(U\) is a nbhd of \(f\of x\) so \(f^{-1}\bof U\) is nbhd of \(x\). Thus \(f^{-1}\bof U\) is open in \(X\). □
Let \(X\) and \(Y\) be topological spaces with
\[ X=\bigcup _{i=1}^{n}X_{i}, \]
where each \(X_{i}\) is open in \(X\). If \(f_{i}:X_{i}\to Y\) is continuous for each \(i=1,2,\ds ,n\) and
\[ f:=\bigcup _{i=1}^{n}f_{i} \]
is a function, then \(f:X\to Y\) is continuous. The same conclusion holds if we assume that each \(X_{i}\) is closed in \(X\).
Proof. Let \(U\) be open in \(Y\). Then
\[ f^{-1}\bof U=\bigcup _{i=1}^{n}f_{i}^{-1}\bof U \]
and \(f_{i}^{-1}\bof U\) is open in \(X_{i}\) for each \(i=1,2,\ds ,n\). Since \(X_{i}\) is open in \(X\), it follows that \(f_{i}^{-1}\bof U\) is open in \(X\) for each \(i=1,2,\ds ,n\). Thus \(f^{-1}\bof U\) is open in \(X\).
Assuming that each \(X_{i}\) is closed in \(X\), we use a similar argument starting with a closed subset of \(Y\). □
Let \(X\) and \(Y\) be topological spaces with \(X=\bigcup _{i\in A}X_{i}\), where each \(X_{i}\) is open in \(X\). If \(f_{i}:X_{i}\to Y\) is continuous for each \(i\in A\) and \(f:=\bigcup _{i\in A}f_{i}\) is a function, then \(f:X\to Y\) is continuous.
Let \(X:=\br \) with \(X_{r}:=\set r\) for each \(r\in \br \). Then each \(X_{r}\) is closed in \(X\). If \(f_{r}:X_{r}\to \br \) is defined by \(f_{r}\of r:=1\) for \(r\in \bq \) and \(f_{r}\of r:=0\) for \(r\in \br \sem \bq \), then \(f_{r}\) is continuous for each \(r\in \br \), but the functions \(f:=\bigcup _{i\in A}f_{i}\) is not continuous.
Let \(X\) be a topological space and \(\sa \) be a family of subsets of \(X\). We say that \(\sa \) is locally finite when each \(x\in X\) has a nbhd \(U\) such that
\[ \set {A\in \sa :A\cap U\neq \emp } \]
is finite.
Let \(X\) be a topological space and \(\sa \) be a locally finite family of subsets of \(X\). Then \(\bigcup _{A\in \sa }\ob A\) is closed.
Proof. Let
\[ x\in \ob {\bigcup _{A\in \sa }\ob A}. \]
There is an open nbhd \(U\) of \(x\) such that
\[ \sa ’:=\set {A\in \sa :A\cap U\neq \emp } \]
is finite. Then \(U\cap \ob A=\emp \) for any \(A\in \sa \sem \sa ’\). Suppose, for a contradiction, that
\[ x\nin \bigcup _{A\in \sa }\ob A. \]
For each \(A\in \sa ’\), let \(U_{A}:=X\sem \ob A\). Then \(U_{A}\) is an open nbhd of \(x\) with \(U_{A}\cap \ob A=\emp \). If
\[ V:=U\cap \bigcap _{A\in \sa ’}U_{A}, \]
then \(V\) is a nbhd of \(x\) such that
\[ V\cap \bigcup _{A\in \sa }\ob A=\emp , \]
which is a contradiction. □
In particular, the union of a locally finite family of closed sets is closed.
Let \(X\) and \(Y\) be topological spaces with \(X=\bigcup _{i\in A}X_{i}\), where each \(X_{i}\) is closed in \(X\) and \(\set {X_{i}:i\in A}\) is locally finite. If \(f_{i}:X_{i}\to Y\) is continuous for each \(i\in A\) and \(f:=\bigcup _{i\in A}f_{i}\) is a function, then \(f:X\to Y\) is continuous.
Proof. Let \(C\) be closed in \(Y\). Then \(f_{i}^{-1}\bof C\) is closed in \(X_{i}\) for each \(i\in A\) so it is closed in \(X\). Since \(\set {X_{i}:i\in A}\) is locally finite, it follows that \(\set {f_{i}^{-1}\bof C:i\in A}\) is locally finite. Since
\[ f^{-1}\bof C=\bigcup _{i\in A}f_{i}^{-1}\bof C, \]
it follows that \(f^{-1}\bof C\) is closed. □
Let \(X\) and \(Y\) be topological space. If \(f:X\to Y\) is a bijection such that both \(f\) and \(f^{-1}\) are continuous, then \(f\) is a homeomorphism. If there exists a homeomorphism \(X\to Y\), then we say that \(X\) and \(Y\) are homeomorphic.
The Euclidean space \(\br ^{n}\) is homeomorphic to the open ball \(B\of {0,1}\) in \(\br ^{n}\). The map \(f:\br ^{n}\to B\of {0,1}\) defined by
\[ f\of x:=\frac {x}{1+\nm x} \]
is a homeomorphism.
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). We say that \(f\) is open if \(f\bof U\) is open in \(Y\) for every open \(U\) in \(X\). We say that \(f\) is closed if \(f\bof C\) is closed in \(Y\) for every closed \(C\) in \(X\).
The inclusion function \(f:\cur {0,1}\to \br \) is continuous and open, but not closed.
The inclusion function \(f:\bra {0,1}\to \br \) is continuous and closed, but not open.
The function \(f:\clo {0,2\pi }\to \br ^{2}\) defined by \(f\of x=\ang {\cos x,\sin x}\) is continuous, but it is neither open nor closed.
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\) be a bijection. The following conditions are equivalent:
1. \(f\) is a homeomorphism.
2. \(f\) is continuous and open.
3. \(f\) is continuous and closed.
Proof. Let \(g:=f^{-1}\). Assume 1. Then \(g:Y\to X\) is continuous. If \(U\) is open in \(X\), then \(f\bof U=g^{-1}\bof U\) is open in \(X\). Thus 2. hold. Similarly, we show that 3. holds.
Now assume that 2. holds. If \(U\) is open in \(X\), then \(g^{-1}\bof U=f\bof U\) is open in \(Y\) so 1. holds. Similarly, we show that 3. implies 1. □
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). Then \(f\) is closed if and only if \(\ob {f\bof A}\sub f\bof {\ob A}\) for every \(A\sub X\).
Proof. Assume that \(f\) is closed. Let \(A\sub X\). Since \(f\bof {\ob A}\) is closed and contains \(f\bof A\), it follows that \(\ob {f\bof A}\sub f\bof {\ob A}\).
Now assume that \(\ob {f\bof A}\sub f\bof {\ob A}\) for every \(A\sub X\). Let \(C\) be closed in \(X\). Then
\[ \ob {f\bof C}\sub f\bof {\ob C}=f\bof C. \]
Since \(f\bof C\sub \ob {f\bof C}\), it follows that equality holds so \(f\bof C\) is closed. □
A function \(f:X\to Y\) is continuous and closed if and only if \(f\bof {\ob A}=\ob {f\bof A}\) for any \(A\sub X\).
Let \(X\) and \(Y\) be topological spaces with \(\sb \) being a basis for the topology on \(X\) and \(f:X\to Y\). Then \(f\) is open if and only if \(f\bof B\) is open in \(Y\) for every \(B\in \sb \).
Proof. Assume that \(f\) is open. Since the members of \(\sb \) are open, the image \(f\bof B\) is open in \(Y\) for every \(B\in \sb \).
Now assume that \(f\bof B\) is open in \(Y\) for every \(B\in \sb \). Let \(U\) be open in \(X\). Then \(U=\bigcup \sa \) for some \(\sa \sub \sb \). Since
\[ f\bof U=\bigcup _{A\in \sa }f\bof A \]
and \(f\bof A\) is open in \(Y\) for every \(A\in \sa \), it follows that \(f\bof U\) is open in \(Y\). Thus \(f\) is open. □
Let \(X\) and \(Y\) be topological spaces and \(f:X\to Y\). The function \(f\) is a topological embedding if it is a homeomorphism onto \(f\bof X\).
\(f\) is a topological embedding provided it is injective, continuous and it’s inverse as a function \(f\bof X\to X\) is continuous.
The function \(f:\clo {0,2\pi }\to \br ^{2}\) defined by \(f\of x=\ang {\cos x,\sin x}\) is injective and continuous, but it is not a topological embedding.
Let \(X\) be an uncountable set with the cofinite (or cocountable) topology. Show that every continuous function \(X\to \br \) is constant.
Give an example of topological spaces \(X\), \(Y\) a function \(f:X\to Y\) and a subspace \(A\sub X\) such that \(f\res A\) is continuous, although \(f\) is not continuous at any point of \(A\).
Let \(X\) be a partially ordered set. Define a topology on \(X\) be declaring \(U\sub X\) to be open if it satisfies the condition: if \(y\le x\) and \(x\in U\), then \(y\in U\). Show that a function \(f:X\to X\) is continuous if and only if it is order preserving (i.e., \(x\le x’\) implies that \(f\of x\le f\of {x’}\)).
Prove that the addition function \(\br ^{2}\to \br \) is open, but is not closed.