constant functions and continuity

It is easy to see that every constant function between topologicalspaces is continuous. A converse result is as follows.

Theorem.

Suppose $X$ is path connected and $D$ is a countablediscrete topological space. If $f\mathrm{:}X\mathrm{\to }D$ is continuous,then $f$ is a constant function.

Proof.

By this result (http://planetmath.org/FiniteAndCountableDiscreteSpaces)we can assume that $D$ is either $\{1,\mathrm{\dots },n\}$, $n\ge 2$or $\mathbb{Z}$, and these are equipped with the subspace topology of $ℝ$.Suppose $f(X)$ has at least two distinct elements, say$\alpha ,\beta \in \mathbb{Z}$ so that

$$f(x)=\alpha ,f(y)=\beta$$

for some $x,y\in X$. Since $X$ is path connected there is a continuous path$\gamma :[0,1]\to X$ such that $\gamma (0)=x$ and $\gamma (1)=y$.Then $f\circ \gamma :[0,1]\to D$ is continuous.Since $D$ has the subspace topology of $ℝ$,this result (http://planetmath.org/ContinuityIsPreservedWhenCodomainIsExtended)implies thatalso $f\circ \gamma :[0,1]\to ℝ$ is continuous.Since $f\circ \gamma$achieves two different values, it achieves uncountably many values,by the intermediate value theorem.This is a contradiction since $f\circ \gamma ([0,1])$is countable.∎