Does a continuous function preserve metrizability

continuitygeneral-topologymetrizabilityproof-verification

This is problem 6.1.8 of S. Morris's "Topology without Tears":

Let $f$ be a continuous mapping of a metrizable space $(X,\tau)$ onto a topological space $(Y,\tau_1)$. Is $(Y, \tau_1)$ necessarily metrizable?

My Solution (by Counterexample):

Let $X$ be the Real line with the standard Euclidean topology. Clearly, $f$ is metrizable.

Let $Y = \{0,1\}$ with the indiscrete topology, i.e., $\tau_1 = \{\emptyset, \{0,1\}\}$. An indiscrete space with at least two points is not metrizable

Define $f$ as follows:

$
f(x)=
\begin{cases}
0&\text{if}\, x\in \mathbb{Q}\\
1&\text{if}\, x\in \mathbb{I}
\end{cases}
$

So, every open interval in $\mathbb{R}$ maps to $\{0,1\}$, and the inverse image of $\{0,1\}$ is $\mathbb{R}$. So, $f$ maps open sets to open sets.

Therefore, $f$ is a continuous function from a metrizable space to a non-metrizable space.

Best Answer

Your use of the word discrete in "Let $Y=\{0,1\}$ with the discrete topology" is the opposite of what you mean.

Also, "So, $f$ maps open sets to open sets." is not what defines the continuity of $f$. You have to check that $f^{-1}$ maps open sets to open sets (which you did).

All in all, the counterexample is correct.

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