Every map from a space with the discrete topology is continuous

Statement

Given two topologies, [ilmath](X,\mathcal{P}(X))[/ilmath] and [ilmath](Y,\mathcal{J})[/ilmath] where:

• [ilmath](X,\mathcal{P}(X))[/ilmath] denotes the discrete topology on [ilmath]X[/ilmath]

We have the following:

• Any mapping, [ilmath]f:X\rightarrow Y[/ilmath], is continuous

Proof of claim

Recall there are two definitions of continuity, the topological:

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

Then there's the metric space definition:

• [ilmath]f:(X,d_1)\rightarrow(Y,d_2)[/ilmath] is continuous if [ilmath]\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][/ilmath]

Topological sense of continuity

Proof:

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

Let [ilmath]A\in\mathcal{J} [/ilmath] be given.

Then [ilmath]f^{-1}(A)\subseteq X[/ilmath] by definition of pre-image

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

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

As [ilmath]A[/ilmath] was arbitrary, this completes the proof.

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

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

Metric sense of continuity

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

• [ilmath]\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][/ilmath]

Proof:

Let [ilmath]x\in X[/ilmath] be given

Let [ilmath]\epsilon > 0[/ilmath] be given

Choose [ilmath]\delta=\tfrac{1}{2}[/ilmath]

Let [ilmath]y\in X[/ilmath] be given

If [ilmath]d(x,y)<\tfrac{1}{2}[/ilmath] we must have [ilmath]x=y[/ilmath]

• (As [ilmath]x\ne y\implies d(x,y)=1[/ilmath] so by contrapositive we have [ilmath]d(x,y)\ne 1\implies x=y[/ilmath])

As [ilmath]x=y[/ilmath] we see [ilmath]d(x,y)=0[/ilmath] and [ilmath]0<\delta[/ilmath]

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

By definition of a metric [ilmath]f(x)=f(y)\implies d'(f(x),f(y))=0[/ilmath]

[ilmath]d'(f(x),f(y))=0<\epsilon[/ilmath] so [ilmath]d'(f(x),f(y))<\epsilon[/ilmath]

As [ilmath]x[/ilmath] was arbitrary we have shown this is true for all [ilmath]x\in X[/ilmath]

That completes the proof.