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I know that every metric space is a topological space. However, not all topological spaces are metric spaces, for example the cofinite topology is not metric since it is not Hausdorff.

But, I still have question which is whether every topological space is the continuous image of a metric space.

I am thinking about considering a discrete space in the domain to ensure that the function is continuous. But I could not finish the argument. Any help will be appreciated.

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  • $\begingroup$ Right. Every subset of a discrete space is an open set; so every function on a discrete space is continuous. (Since the inverse of any open set is open). $\endgroup$ – user4894 Aug 13 '16 at 19:26
  • $\begingroup$ If you detrivialize the question by moving from continuous images to quotient maps, then you obtain the class of sequential spaces. $\endgroup$ – user87690 Aug 14 '16 at 14:17
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Yes, your idea in the second paragraph is correct. Suppose $(X, \tau)$ is any topological space, and $A$ is any set at least as big (in the sense of cardinality) as $X$. (For instance, take $A=X$.)

Endow $A$ with the discrete metric: $d(a, b)=1$ for $a\not=b$. Then any map from $A$ to $X$ is continuous (exercise); since $A$ is at least as big as $X$, there exists a surjection $f$ from $A$ to $X$.

The exercise, by the way, is maybe better expressed as:

If $f:U\rightarrow V$ is a map between topological spaces, and $U$ is discrete, then $f$ is continuous.

HINT: $f$ is continuous iff the preimage of every $V$-open set is $U$-open. If $U$ is discrete, are there any subsets of $U$ which are not open?

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  • $\begingroup$ No, $U$-discrete eash subset will open and closed at the same time.By the way, thank you for clarification. $\endgroup$ – Gob Aug 13 '16 at 20:34

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