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As a part of self study, I am trying to prove the following statement:

Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$.

Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?

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7 Answers 7

up vote 13 down vote accepted

Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:

If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.

Now using our closure property for $D$: $$f[\overline{D}] \subset \overline{f[D]} = \overline{f[f^{-1}[C]]} \subset \overline{C} = C,$$ as $C$ is closed.

This means that $\overline{D} \subset f^{-1}[C] = D$, making $D$ closed, as required.

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I agree, no need to mess with complements. –  wildildildlife Feb 28 '12 at 22:06
    
$C$ should be closed in $Y$, not $X$. –  Holdsworth88 Feb 29 '12 at 10:16
    
@Pjennings thanks, corrected! –  Henno Brandsma Feb 29 '12 at 16:59
    
There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c \text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(\overline{D})\subseteq \overline{f(D)}\text{???}$ –  Freeze_S Jan 22 at 16:57
    
@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here. –  Carsten Schultz Jan 22 at 18:09

If $f$ is continuous, then $f^{-1}(Y-\overline{f(A)})$ is open; since $f^{-1}(Y-\overline{f(A)}) = X -f^{-1}(\overline{f(A)})$, then $f^{-1}(\overline{f(A)})$ is closed; since $A\subseteq f^{-1}(f(A))\subseteq f^{-1}\overline{f(A)}$, we conclude that $\overline{A}\subseteq f^{-1}(\overline{f(A)})$, and therefore $f\left(\overline{A}\right)\subseteq f\left(f^{-1}\left(\overline{f(A)}\right)\right)\subseteq \overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).

Conversely, assume that for every $A$, $f\left(\overline{A}\right)\subseteq \overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=\overline{X-f^{-1}(V)}$.

By assumption, $f\left(\overline{X-f^{-1}(V)}\right)\subseteq \overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) \subseteq Y-V$, which is closed; so $$f\left(\overline{X-f^{-1}(V)}\right)=f\left(\overline{f^{-1}(Y-V)}\right)\subseteq\overline{f\left(f^{-1}(Y-V)\right)}\subseteq Y-V.$$ Therefore, $$\overline{X-f^{-1}(V)} \subseteq f^{-1}(Y-V).$$ Is this sufficient?

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That was more than sufficient, thank you. –  Holdsworth88 Feb 28 '12 at 16:51

We show that $f$ is continuous at each $x\in X$. Recall that $f$ is continuous at $x$ if for any open nhood $V$ of $f(x)$, there is an open nhood $U$ of $x$ with $f(U)\subseteq V$.

So, let $x\in X$ and let $V$ be an open nhood of $f(x)$. Set $E=X\setminus f^{-1}(V)$. Noting that $f(E)$ is contained in the closed set $V^C$, we see that $$f(\overline{E})\subseteq \overline{f(E)}\subseteq V^C; $$ whence $x\notin \overline E$ (since $f(x)$ is in $V$).

But then $x$ is in the open set $X\setminus \overline{E}$. Moreover, since $X\setminus \overline{E}\subseteq X\setminus E=f^{-1}(V)$, it follows that $f(X\setminus \overline{E})\subseteq V$, as desired.




An aside:

If $X$ were first countable, we could argue using sequences (where a sequence $(x_n)$ converges to $x$ if for any nhood $U$ of $x$ there is an $N$ so that $x_m\in U$ for all $m\ge N$); for in such a space a function is continuous at $x$ if and only if $(f (x_n))$ converges to $f(x)$ whenever $x_n$ converges to $x$.

Indeed, it is easily verified that given $x_n\rightarrow x$ and any subsequence $(x_{n_k})$ of $(x_n)$, that the image of this subsequence under $f$ when thought of as a sequence has a subsequence that converges to $f(x)$. Thus every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$, which implies that $(f(x_n))$ converges to $x$.

Of course, $X$ need not be first countable...

Though I think the sequential method above is a bit much here, I wonder if the argument can be suitably modified to arbitrary $X$ by using nets?

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Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.

Suppose that $f$ is not continuous. Then there is some point $x_0\in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_V\in V$ such that $f(x_V)\notin U$. Let $$A=\{x_V:V\text{ is an open nbhd of }x_0\}\;.$$

Clearly $x_0\in\overline{A}$, and $f[A]\subseteq Y\setminus U$. Moreover, $Y\setminus U$ is closed, so $\overline{f[A]}\subseteq Y\setminus U$, and hence

$$y_0\in f[\overline{A}]\subseteq\overline{f[A]}\subseteq Y\setminus U\;,$$

contradicting the choice of $U$.

Those who like nets may notice that $A$ actually is a net, over the directed set $\langle\mathscr{N}(x_0),\supseteq\rangle$, where $\mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)

Those who prefer filters may modify this to consider the filter $$\mathscr{F}=\{V\cap f^{-1}[Y\setminus U]:V\in\mathscr{N}(x_0)\}$$ instead.

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Here's one proof of the converse provided $X$ and $Y$ are metric spaces:

Take a limit point $x$ of $A$. Then because $f(\overline{A}) \subseteq \overline{f(A)}$, we have that $f(x)$ is a limit point of $f(A)$.

Because $x$ is a limit point of $A$, for every $\delta > 0$ there is a point $p \in A$ with $|x - p| < \delta$. And because $f(x)$ is a limit point of $f(A)$, for every $\epsilon > 0$, there is a point $q \in f(A)$ with $|f(x) - f(p)| < \epsilon$.

We thus have that for every point $x \in A$ with $|x - p| < \delta$, in fact $|f(x) - f(p)| < \epsilon$, so $f$ must be continuous.

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2  
I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$. –  kahen Feb 28 '12 at 16:45
2  
We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates. –  user21436 Feb 28 '12 at 16:46
    
Can't you just substitute $d(x, p)$ for $|x - p|$? –  jamaicanworm Feb 28 '12 at 16:50
3  
@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable. –  Arturo Magidin Feb 28 '12 at 16:50
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@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia. –  Arturo Magidin Feb 28 '12 at 16:59

A proof using nets:

Suppose $y\in f(\overline{A})\backslash \overline{fA}$. Taking $x\in \overline{A}$ with $y=f(x)$; there is a net $x_i\in A$ converging to $x$. Now $f(x_i)$ is a net in $f(A)$ which does not converge to $f(x)$ [for this would imply $f(x)\in \overline{fA}$]. Thus $f$ is not continuous.

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This is the direction that the OP had already been able to show. –  Brian M. Scott Feb 29 '12 at 6:08
    
@Brian: Yes, sorry! I realized this the day after... –  wildildildlife Mar 2 '12 at 13:15

The assertion is equivalent to:
$\overline{A}\subseteq f^{-1}(\overline{f(A)})$
So, the assertion follows from:
$\overline{A}\subseteq\overline{f^{-1}(f(A))}\subseteq\overline{f^{-1}(\overline{f(A)})}=f^{-1}(\overline{f(A)})$

  1. Inclusion: $A\subseteq f^{-1}(f(A)) \Rightarrow \overline{A}\subseteq\overline{f^{-1}(f(A))}$
  2. Inclusion: $f(A)\subseteq\overline{f(A)} \Rightarrow f^{-1}(f(A))\subseteq f^{-1}(\overline{f(A)}) \Rightarrow \overline{f^{-1}(f(A))}\subseteq \overline{f^{-1}(\overline{f(A)})}$
  3. Equality: $\overline{f(A)} \text{ closed} \Rightarrow f^{-1}(\overline{f(A)}) \text{ closed} \Rightarrow \overline{f^{-1}(\overline{f(A)})}=f^{-1}(\overline{f(A)})$

The converse assertion is equivalent to:
$\overline{B}=B \Rightarrow \overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)\subseteq\overline{f^{-1}(B)}\subseteq f^{-1}(f(\overline{f^{-1}(B)}))\subseteq f^{-1}(\overline{f(f^{-1}(B))}) \subseteq f^{-1}(\overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=\overline{f^{-1}(B)}$

  1. Inclusion: $A\subseteq \overline{A} \text{ in general}$
  2. Inclusion: $A\subseteq f^{-1}(f(A)) \text{ in general}$
  3. Inclusion: $f(\overline{A})\subseteq \overline{f(A)} \text{ by assumption}$
  4. Inclusion: $f(f^{-1}(B))\subseteq B \text{ in general} \Rightarrow \overline{f(f^{-1}(B))}\subseteq \overline{B} \Rightarrow f^{-1}(\overline{f(f^{-1}(B))})\subseteq f^{-1}(\overline{B})$
  5. Equality: $\overline{B}=B \Rightarrow f^{-1}(\overline{B})=f^{-1}(B)$
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I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer. –  Pedro Tamaroff Jan 22 at 17:51
    
I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer. –  Freeze_S Jan 22 at 17:57

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