Every map from a space with the discrete topology is continuous

Statement

Given two topologies, $(X,\mathcal{P}(X))$ and $(Y,\mathcal{J})$ where:

• $(X,\mathcal{P}(X))$ denotes the discrete topology on $X$

We have the following:

• Any mapping, $f:X\rightarrow Y$, is continuous

Proof of claim

Recall there are two definitions of continuity, the topological:

• $f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$ is continuous if $\forall A\in\mathcal{K}[f^{-1}(A)\in\mathcal{J}]$ - for all open sets in $Y$ the pre-image under $f$ is open in $X$

Then there's the metric space definition:

• $f:(X,d_1)\rightarrow(Y,d_2)$ is continuous if $\forall x\in X\forall\epsilon\in\mathbb{R}_{>0}\exists\delta\in\mathbb{R}_{>0}\forall y\in X[d_1(x,y)<\delta\implies d_1(f(x),f(y))<\epsilon]$

Topological sense of continuity

Proof:

• We wish to prove: $\forall A\in\mathcal{J}[f^{-1}(A)\in\mathcal{P}(X)]$

Let $A\in\mathcal{J}$ be given.

Then $f^{-1}(A)\subseteq X$ by definition of pre-image

• Which recall is: $f^{-1}(A):=\{x\in X\vert f(x)\in A\}$

As every subset of $X$ is open in $X$ we see that $f^{-1}(A)$ is open.

As $A$ was arbitrary, this completes the proof.

• This isn't surprising as by definition $A\subseteq X\iff A\in\mathcal{P}(X)$ so we really wanted to show $\forall A\in\mathcal{J}[f^{-1}(A)\subseteq X]$

There is no need to proceed and consider the metric space, however it is included for completeness.

Metric sense of continuity

Suppose that $(X,d)$ is a metric space where $d$ denotes the discrete metric and $(Y,d')$ is any metric space.
We wish to show:

• $\forall x\in X\forall\epsilon\in\mathbb{R}_{>0}\exists\delta\in\mathbb{R}_{>0}\forall y\in X[d(x,y)<\delta\implies d'(f(x),f(y))<\epsilon]$

Proof:

Let $x\in X$ be given

Let $\epsilon > 0$ be given

Choose $\delta=\tfrac{1}{2}$

Let $y\in X$ be given

If $d(x,y)<\tfrac{1}{2}$ we must have $x=y$

• (As $x\ne y\implies d(x,y)=1$ so by contrapositive we have $d(x,y)\ne 1\implies x=y$)

As $x=y$ we see $d(x,y)=0$ and $0<\delta$

y definition of a function $x=y\implies f(x)=f(y)$ (a function must map a point to exactly one point of the image - it cannot map to two things)

By definition of a metric $f(x)=f(y)\implies d'(f(x),f(y))=0$

$d'(f(x),f(y))=0<\epsilon$ so $d'(f(x),f(y))<\epsilon$

As $x$ was arbitrary we have shown this is true for all $x\in X$

That completes the proof.