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

I am trying to show that given $f:M \rightarrow N$, where $M$ is compact, $f$ is continuous and onto, then given $A \subset N$:

$$ f^{-1}(A) \text{ closed} \implies A\text{ closed} $$

I am dealing here with any metric space, although I feel that the approach is identical to the $\mathbb{R}$eal case (am I right?). So my attempt is the following.

My attempt: $f^{-1}(A)$ closed and $f^{-1}(A) \subset M \implies f^{-1}(A)$ bounded and hence, compact. Hence, I can consider $\{x_n\} \rightarrow x, x \in f^{-1}(A)$. Using continuity, now we have a $\{[f(x)]_n\} \subset N$ and $\{[f(x)]_n\} \rightarrow f(x)$. Since $x \in f^{-1}(A)$, $f(x)=y \in A$. But how do I know that all limit points of $A$ are generated by convergent subsequences in $f^{-1}(A)$?


share|cite|improve this question
What is a metric space in your question? $M$? $N$? Both? – Olivier Bégassat Feb 2 '14 at 19:20
Both of them. Actually the question does not specify. – user191919 Feb 2 '14 at 19:23
up vote 0 down vote accepted

Let $\{y_n\}_{n\in\mathbb N}\subset A$, with $y_n\to y\in N$. We need to show that $y\in A$.

Since $f$ is onto, there exist a sequence $\{x_n\}_{n\in\mathbb N}\subset f^{-1}(A)\subset M$, with $f(x_n)=y_n$.

But as $M$ is compact, there exists a converging subsequence $x_{n_k}\to x\in M$. By the hypothesis, $f^{-1}(A)$ is closed, and thus $x\in f^{-1}(A)$.

We know that $y_{n_k}=f(x_{n_k})\to f(x)$, and as the limit of $\{y_n\}_{n\in\mathbb N}$ is unique, we have that $y=f(x)$.

However, $x\in f^{-1}(A)$ implies that $y=f(x)\in f\big(f^{-1}(A)\big)=A$.

The last one, i.e., $f\big(f^{-1}(A)\big)=A$, is a property of the surjections. (Onto maps.)

Thus $A$ is indeed closed.

share|cite|improve this answer

My solution assumes $N$ is Hausdorff space (metric spaces would have this property, for example).

As $f$ is onto, we have $f(f^{-1}(A)) = A.$ Now $f$ is continuous and $f^{-1}(A)$ is compact (closed set in a compact space), then $f(f^{-1}(A)) = A$ must be compact, hence closed (compact set in a Hausdorff space).

(Not relevant for this solution, but note also that $N$ is a compact space, as $f$ is continuous and onto, and $M$ is compact.)

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.