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I have two questions: 1.) I have been thinking a while about the fact, that in general the union of closed sets will not be closed, but I could not find a counterexample, does anybody of you have one available?

2.) The other one is, that I thought that one could possibly say that a function f is continuous iff we have $f(\overline{M})=\overline{f(M)}$(In the second part this should mean the closure of $f(M)$). Is this true?

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    $\begingroup$ 1) take $\bigcup_{n\in\mathbb{N}}[1/n,1]$ $\endgroup$ – user127.0.0.1 Jun 29 '13 at 9:57
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    $\begingroup$ The second statement is missing a quantifier (for all subsets $M$ of $X$?). And are you defining $f$ to be closed as implying also that it is continuous? $\endgroup$ – Henno Brandsma Jun 29 '13 at 10:10
  • $\begingroup$ yes for all subsets and no f should not be closed $\endgroup$ – user66906 Jun 29 '13 at 10:14
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(1) Take the closed sets

$$\left\{\;C_n:=\left[0\,,\,1-\frac1n\right]\;\right\}_{n\in\Bbb N}\implies \bigcup_{n\in\Bbb N}C_n=[0,1)$$

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Or to be even lazier, consider that generally singletons are closed (for varying definitions of "generally"). A set is always the union of its points, so you can write $(0,1)$ as $\bigcup_{x \in (0,1)} \{ x \}$.

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(2) is not true. Because it implies that $f$ is a closed map, i.e, maps closed sets to closed sets. However, there are continuous maps which are not closed. For example consider the projection $\pi: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $(x,y) \mapsto x$. Observe that the set $$C = \{(x,1/x) : x\in \mathbb{R} - \{0\}\}$$ is closed in $\mathbb{R}^2$. However, $\pi(C) = \mathbb{R} - \{0\}$ is not closed in $\mathbb{R}$. Yet $\pi$ is definitely continuous.

The correct equivalence for continuity given by the closure operator is $f(\overline{M}) \subseteq \overline{f(M)}$ for every subset $M$ of $X$.

Edit: Here's my argument for why $C$ is closed. Suppose $(a,b) \in \mathbb{R}^2$ lies inside $\overline{C}$. We will show that $(a,b)$ must be in $C$. Since $\mathbb{R}^2$ is first countable, there exists a sequence $y_n$ in $C$ converging to $(a,b)$. By the definition of $C$, we can write $y_n = (x_n,1/x_n)$. Therefore $x_n$ converges to $a$ and $1/x_n$ converges to $b$. Now note that if $x_n$ were to converge to $0$, $1/x_n$ would diverge. Hence we conclude that $a \neq 0$. Thus $1/x_n$ converges to $1/a$. Since we are in a Hausdorff space, the sequence $1/x_n$ cannot converge to two different points. Thus $b = 1/a$ and hence $(a,b) = (a,1/a) \in C$.

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    $\begingroup$ and it sounds so nice: "close points map to close points" $\endgroup$ – citedcorpse Jun 29 '13 at 10:33
  • $\begingroup$ $C:=\{(x, 1/x): x\in \mathbb{R} -{0}\}$ is closed? I do not see it. It is unbounded right? $\endgroup$ – user51196 Jun 29 '13 at 10:56
  • $\begingroup$ Sorry, I guess I have misunderstood Compact set (closed and bounded), with a Closed set. However, I do not see why your set C is closed. $\endgroup$ – user51196 Jun 29 '13 at 11:06
  • $\begingroup$ @noether it's the graph of $1/x$, which is continuous where defined, so it's a closed set (intuitively it's clear that the complement is open) $\endgroup$ – citedcorpse Jun 29 '13 at 11:15
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    $\begingroup$ @noether: I included an argument for why $C$ is closed in my edit. Let me know if something's wrong with it. $\endgroup$ – Cihan Jun 29 '13 at 21:15
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A function $f:X \to Y$ between topological spaces sends closed sets to closed sets (what I call $f$ is closed) iff

$$ \forall A \subset X: \overline{f[A]} \subset f[\overline{A}]$$

If $f$ is closed and $A \subset X$, then $\overline{A}$ is closed, so $f[\overline{A}]$ is also closed. As $A \subset \overline{A}$, $f[A] \subset f[\overline{A}]$ and so also $\overline{f[A]} \subset \overline{f[\overline{A}]} = f[\overline{A}]$ as the latter set is closed.

On the other hand, if $f$ satisfies the closure property, and $C \subset X$ is closed, then $$ f[C] \subset \overline{f[C]} \subset f[\overline{C}] = f[C]$$

as $C$ is closed. It follows that $f[C]$ equals its closure, hence is closed. So $f$ is a closed map.

Sort of dually: $f$ is continuous iff

$$\forall A \subset X : f[\overline{A}] \subset \overline{f[A]}$$

So the other inclusion then holds.

If $f$ is continuous, and $A \subset X$, $$A \subset f^{-1}[f[A]] \subset f^{-1}[\overline{f[A]}]$$

and the latter set is closed by continuity of $f$ (inverse images of closed sets are closed). So as it contains $A$, it also contains $\overline{A}$, which is the smallest closed set containing $A$. So

$$ \overline{A} \subset f^{-1}[\overline{f[A]}]$$

which implies $f[\overline{A}] \subset \overline{f[A]}$. On the other hand, if $f$ satisfies the second closure property, let $C \subset Y$ be closed. Then taking $A = f^{-1}[C]$ then

$$f[\overline{f^{-1}[C]}] \subset \overline{f[f^{-1}[C]]} \subset \overline{C} = C$$

This implies (by definition of $f^{-1}$) that $\overline{f^{-1}[C]} \subset f^{-1}[C]$ which means that $f^{-1}[C]$ is closed. So $f$ is continuous, as inverse images of closed sets are closed.

So the equality $\overline{f[A]} = f[\overline{A}]$ for all subsets $A$ of $X$ is exactly saying that $f$ is both closed and continuous.

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