Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question "Is $f(X)$ compact?" is something that occured to me when attempting the Munkres question.

I think $f(X)$ is compact. Let $ \{V_\alpha \}$ be an arbitrary open cover of $Y$ such that $ \bigcup_\alpha V_\alpha \supset Y \supset f(X).$ $\{V_\alpha \}$ is an open cover of $f(X)$.

Since $Y$ is compact, then there exists a finite subcover such that $ f(X) \subset Y \subset \bigcup_{i=1}^n V_{\alpha_i}$.

Lemma 26.1 of Munkres : Let $Y$ be a subspace of $X$. Then $Y$ is compact iff every covering of $Y$ by sets open in $X$ contains a finite subcollection covering $Y$.

So by the above lemma, $f(X)$ is compact as a subspace of $Y$.

I am not sure if it is right though, please kindly point out the mistakes.

Thank you.

share|cite|improve this question
You have to consider an arbitrary cover of $f(X)$. Your $(V_\alpha)$ is not random enough as you require it to be also a cover of $Y$. – Stefan Hamcke Mar 2 '13 at 21:48
Furthermore, this would imply that any subset of a compact space is compact, since it can be expressed as the continuous image of the inclusion of this subset. – Stefan Hamcke Mar 2 '13 at 21:50
up vote 10 down vote accepted

It suffices to find a non-compact subset $X$ of a compact Hausdorff space $Y$, because the inclusion map $i:X\hookrightarrow Y$ is always continuous. An easy one to consider is $X=(0,1)$ and $Y=[0,1]$.

share|cite|improve this answer

Let $X$ be $\Bbb Z^+$ with the discrete topology, let $Y=[0,1]$ with the usual topology, and let $$f:X\to Y:n\mapsto\frac1n\;.$$ Then $f$ is continuous, $Y$ is compact Hausdorff, and $f[X]$ is not even closed in $Y$, let alone compact. A very simple open cover of $f[X]$ that has no finite subcover is

$$\left\{\left(\frac1{n+1},\frac1{n-1}\right):n\ge 2\right\}\cup\left\{\left(\frac12,1\right]\right\}\;,$$

and it illustrates what’s wrong with your argument: it covers $f[X]$, but it *doesn’t cover $Y$. It covers only $(0,1]$.

share|cite|improve this answer

Note that the continuity of $f$ is not enough to conclude that $f(X)$ is closed in $Y$. If $f(X)$ is closed then it is compact because a closed subset of a compact set is compact.

For example, fix $\{q_n\mid n\in\Bbb N\}$ an enumeration of $\Bbb Q\cap[0,1]$, then $n\mapsto q_n$ is continuous from the discrete space $\Bbb N$ into $[0,1]$ (every function from a discrete space is continuous), but $\Bbb Q\cap[0,1]$ is certainly not compact nor it is closed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.