0
$\begingroup$

Exercise: Let $f:(X,d)\to (Y,\rho)$ be a continuous function. Let $A\subset X$ such that $\overline{A} = X$. Show that if $f$ is surjective, then $\overline{f(A)} = Y$.

What I've tried: If $f$ is surjective then for every $y\in Y$ there exists at least one $x\in X$ such that $f(x) = y$. We have $f(A) = \{f(x)\text{ for } x\in A\}$, and so $\overline{f(A)} = \overline{\{f(x)\text{ for }x\in A\}}$. I don't really know how to translate this into something useful.

Question: How do I solve this exercise?

$\endgroup$
1
$\begingroup$

Hint:

It can be shown that function $f:X\to Y$ is continuous iff $f(\overline B)\subseteq\overline {f(B)}$ for every $B\subseteq X$.

For a proof of that see here.

$\endgroup$
0
$\begingroup$

Here is a fairly short proof that uses the following description of dense sets: if $A \subset X$ then $\overline A = X$ if and only if every nonempty open set in $X$ intersects $A$.

Let $U \subset Y$ be an open set. Then $f^{-1}(U)$ is open too, and nonempty since $f$ is surjective.

Since $A$ is dense in $X$, there exists a point $x \in A \cap f^{-1}(U)$. Consequently $f(x) \in f(A) \cap U$.

Thus $f(A)$ intersects every open set $U \subset Y$, so that $f(A)$ is dense in $Y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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