2

Let $f : X \longrightarrow Y$ be continuous, and let $A \subseteq X$. Show that $f(\overline A) \subseteq \overline{f(A)}$.

My attempt

Let $y \in f(\overline A)$. Then there exists $x \in \overline A$ such that $f(x) = y$. Now let us take an open neighbourhood $V$ of $y$ in $Y$ arbitrarily. If we can show that $V \cap f(A) \neq \emptyset$ then our purpose will be served.

Now since $f(x) = y \in V$, we have $x \in f^{-1} (V)$. Since $x \in \overline A$ and $f^{-1} (V)$ is open in $X$ (since $f$ is continuous), we have that $f^{-1} (V) \cap A \neq \emptyset$.

Let $z \in f^{-1} (V) \cap A$. Then $f(z) \in V \cap f(A)$, which proves that $V \cap f(A) \neq \emptyset$ and we are done.

Is my reasoning correct at all? Please verify it.

Did
  • 284,245
Arnab Chattopadhyay.
  • 2,698
  • A period must be followed by a space. – Kenny Lau Aug 17 '17 at 08:31
  • @KennyLau Very true. And not every period should be followed by a blank line. – Did Aug 17 '17 at 08:36
  • I think it is correct. – Idonknow Aug 17 '17 at 08:38
  • It is correct. For a slightly different approach (but basically the same) see the second part of this answer: https://math.stackexchange.com/a/2115237/385702 – klirk Aug 17 '17 at 08:44
  • @yanko, ok, I clarify: the proof is correct! $y$ is in the closure of $f(A)$ if every open nbhd of $y$ intersects $f(A)$. This is the definition of a point of closure. Also, it is not assumed that $y\in f(A)$, only $y\in f(\bar A)$. – klirk Aug 17 '17 at 08:48

1 Answers1

2

Your reasoning is correct.

Alternative route:

$f$ is continuous and $\overline{f(A)}$ is closed, so we conclude that $f^{-1}\left(\overline{f(A)}\right)$ is closed.

This with: $$A\subseteq f^{-1}\left(f(A)\right)\subseteq f^{-1}\left(\overline{f(A)}\right)$$allowing us to conclude:$$\overline{A}\subseteq f^{-1}\left(\overline{f(A)}\right)$$ or equivalently: $$f(\overline{A})\subseteq\overline{f(A)}$$

drhab
  • 153,781