Topological Continuity

Definition (Topological Characterization of Continuity)

If $(X,{\tau }_x)$ and $(Y,{\tau }_y)$ are topological spaces, then we can define a function $f:X\to Y$ to be continuous if $\forall$ open $U\subseteq Y$, the preimage $f^{-1}(U)$ is open in $X$. This is correct if the topologies come from metrics.

This is also in Proposition (Topological Characterization of Continuity).

Example 1

1. Any map out of a space with the discrete topology is continuous.
2. Any map into a space with the indiscrete topology is continuous.

Proof: For $(1)$, every subset of $X$ is open. For $(2)$, the topology defined in $Y$ is only $\{\varnothing ,Y\}$, which means the preimage for any of these sets is merely $\varnothing$ and $X$ respectively.

Example 2

If $\beta$ is a basis for ${\tau }_Y$, then it’s enough to check that $f^{-1}(B)$ is open in $X$, $\forall B\in \beta$.

Proof:

$$f^{-1}\left(\bigcup _{i\in I}{B}_i\right)=\bigcup _{i\in I}{f}^{-1}(B_i)$$

will be open.

Definition (Continuity at a Point)

The function $f:(X,{\tau }_X)\to (Y,{\tau }_Y)$ is continuous at $a\in X$ if $\forall$ neighborhood $N$ of $f(a)$ in $Y$, the preimage $f^{-1}(N)$ is neighborhood of $a$ in $X$.

Theorem (Equivalence of Global & Pointwise Continuity)

A function $f:(X,{\tau }_{X)}\to (Y_{},{\tau }_T)$ is globally continuous $⟺$ it is continuous at $a,\forall a\in X$.

Proof. $(⟸)$ Given $V\subseteq Y$ open, WTS $f^{-1}(V)$ is open in $X$. So take $a\in f^{-1}(V)$. Continuity at this point says that since $V$ is a neighborhood of $f(a)$. we have $f^{-1}(V)$ is a neighborhood of $a$. Since $a$ was arbitrary, then $f^{-1}(V)$ is open.

$(⟹)$ Given $N$ a neighborhood of $f(a)$, then it contains some open $V$ where $f(a)\in V$. Then global continuity implies $f^{-1}(V)$ is open, but $a\in f^{-1}(V)\subseteq f^{-1}(N)$ and so $f^{-1}(N)$ is a neighborhood of $a$.

Theorem (Closed Set Criterion for Continuity)

$f$ is continuous $⟺f^{-1}(F)$ is closed in $X$, $\forall$ closed $F$ in $Y$.

Proof: The preimage satisfies $f^{-1}(Y-U)=X-f^{-1}(U)$. Then apply Definition (Topological Characterization of Continuity).

Also in Proposition (Topological Characterization of Continuity) from real analysis.

Theorem (Characterization of Continuity by Closure)

$f$ is continuous $⟺f({\text{Cl}}_X(A))\subseteq {\text{Cl}}_Y(f(A))$ for all subsets $A$ of $Y$.

Proof: $(⟹)$ Given continuous $f$ and a point $x\in \overline{A}$, WTS $f(x)\in \overline{f(A)}$, or that every neighborhood of $f(x)$ meets $f(A)$. But if $N$ is a neighborhood of $f(x)$, then b the local form of continuity, $f^{-1}(N)$ is a neighborhood of $x$.

Therefore, since $x\in \overline{A}$, $f^{-1}(N)$ must meet $A$ at some point $y\in A\cap f^{-1}(N)$. But then $f(y)\in f(A)\cap N$ as requested.

$(⟸)$ We’ll use the “closed-set” definition of continuity. Given $F$ closed in $Y$, show $A=f^{-1}(F)$ closed in $X$. If $x\in \overline{A}$, we’ll show it is also in $A$, such that it implies $\overline{A}=A$. We know

$$f(x)\in \overline{f}(A)=\overline{F}=F$$

hence $x\in f^{-1}(F)=A$ by “condition”.

Theorem (Continuity by Composition in a Topology)

Given continuous functions

$$\begin{array}{rl}f& :(X,{\tau }_X)\to (Y,{\tau }_Y)\\ g& :(Y,{\tau }_Y)\to (Z,{\tau }_Z)\end{array}$$

we have that $g\circ f$ is continuous.

Proof: Given $U$ open in $Z$,

$$\begin{array}{rl}(g\circ f{)}^{-1}(U)& :=f^{-1}(g^{-1}(U))\end{array}$$

But as $g$ continuous, $g^{-1}(U)$ is open, and as $f$ is continuous, $f^{-1}(g^{-1}(U))$ is open. We apply Definition (Topological Characterization of Continuity) twice.