54
$\begingroup$

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?

  • 0
    I am not able to come up with any example of a continuous function s.t. $f(\overline{A}) \subsetneq \overline{f(A)}$. Can anyone give such an examples?2018-03-11
  • 2
    Assuming $f$ is continuous, how exactly is the result "immediate"?2018-05-19

7 Answers 7

54

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 $D$: $$f[\overline{D}] \subseteq \overline{f[D]} = \overline{f[f^{-1}[C]]} \subseteq \overline{C} = C,$$ as $C$ is closed.

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

  • 0
    I agree, no need to mess with complements.2012-02-28
  • 0
    $C$ should be closed in $Y$, not $X$.2012-02-29
  • 0
    There's a misleading in the argumentation:
    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!
    Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(\overline{D})\subseteq \overline{f(D)}\text{???}$
    2014-01-22
  • 1
    @Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.2014-01-22
  • 0
    why the result is imeadiat if $f$ is continuous?2016-05-09
  • 0
    @GuerlandoOCs the OP asked for one direction only.2016-05-09
  • 0
    @HennoBrandsma I didn't know how to solve this question today at my exam, then I closed my eyes and the image of your answer came to my mind. I tried to search an email of yours to tip you $1 in BTC but I couldn't2016-05-25
  • 0
    @mhle a set is closed iff it's equal to its closure.2016-08-23
  • 0
    @HennoBrandsma I meant, how do you jusity that $\bar{D}\subset f^{-1}(C)$?2016-08-23
  • 0
    @mhle $x \in f^{-1}[C]$ iff $f(x) \in C$ and by the previous line this holds for all $x \in \overline{D}$.2016-08-23
22

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?

  • 2
    That was more than sufficient, thank you.2012-02-28
  • 1
    Converse part is nice.2017-01-26
10

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?

10

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.

4

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.

  • 4
    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 $Y$.2012-02-28
  • 3
    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.2012-02-28
  • 0
    Can't you just substitute $d(x, p)$ for $|x - p|$?2012-02-28
  • 4
    @jamaicaworm: Only in a *metric* space. Not every topological space is a metric space, or even metrizable.2012-02-28
  • 0
    I'm not familiar with the definitions of open and closed in a non-metric space. In the future, @Pjennings should post them in questions.2012-02-28
  • 6
    @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](http://en.wikipedia.org/wiki/Topological_space#Definition).2012-02-28
2

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.

  • 0
    This is the direction that the OP had already been able to show.2012-02-29
  • 0
    @Brian: Yes, sorry! I realized this the day after...2012-03-02
1

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: $\overline{B}=B \Rightarrow f^{-1}(\overline{B})=f^{-1}(B)$
  • 1
    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.2014-01-22
  • 0
    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.2014-01-22