Prove that every continuous real-valued function on an indiscrete space is constant, even though such a space is always normal.

Question

Prove that every continuous real-valued function on an indiscrete space is constant, even though such a space is always normal.

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Can I get some help.I could not solve it.

Answer 1

HINT: Let $X$ be a space with the indiscrete topology. If $|X|=1$, then obviously every function on $X$ is constant, so assume that $X$ has at least two points. Suppose that $f:X\to\Bbb R$ is not constant; then there are points $x,y\in X$ such that $f(x)\ne f(y)$. Let $U$ and $V$ be disjoint open neighborhoods of $f(x)$ and $f(y)$, respectively. If $f$ is continuous, what do you know about $f^{-1}[U]$ and $f^{-1}[V]$? Why does this contradict the assumption that $X$ has the indiscrete topology?

Answer 2

Take any value $y$ in the range of $f$. Then $f^{-1}(\{y\})$ is a non-empty closed subset of the domain. What can this be, if the domain is equipped with the trivial (indiscrete if you will) topology?