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Let $f:(X, d_X)\longrightarrow (Y, d_Y)$ be a continuous function between two metric spaces, $A\subseteq X$. We have $f(\overline{A})\subseteq \overline{f(A)}$ from this question. Can you please provide a counter-example to $f(\overline{A})\supseteq \overline{f(A)}$? I.e. in which cases is the forward inclusion $f(\overline{A})\subseteq \overline{f(A)}$ proper?

My thoughts are that $f$ creates a new set of sequences that are not the images $(f(x_n))$ of sequences $(x_n)$ in $X$, with limits in $\overline{f(A)}$ but not in $f(\overline{A})$. I cannot describe those sequences I'm thinking of.

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ahorn
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  • Hint: Show that $f(\overline A)=\overline{f(A)},\forall A\subseteq X$ if and only if $f(F)$ is closed for all closed $F\subseteq X$. – Yai0Phah Apr 08 '16 at 20:33
  • I think this works. Let x = [0,$\infty$) and let f(x) = x if x $\le$ 1; 2 - 1/x if x > 1. Let $A = \mathbb N \cup {1/n| n\in \mathbb N} \cup {0}=\overline A$. Then $f(\overline A) = {0}\cup {1/n| n\in \mathbb N} \cup {2-1/n| n\in \mathbb N} \subsetneq {0}\cup {1/n| n\in \mathbb N} \cup {2-1/n| n\in \mathbb N}\cup {2} = \overline{f(A)}\$. I think that works. – fleablood Apr 08 '16 at 20:55
  • @fleablood Yes, I think that works too. You and Ahmed only considered $A$ closed. I would be interested to see an example with $A$ open. – ahorn Apr 08 '16 at 21:02
  • @fleablood maybe just drop the ${0}$ from $A$, and the example would be stronger. – ahorn Apr 08 '16 at 21:05
  • My initial idea didn't have 0 in A but, to be honest, I didn't want to deal with the LaTex of retyping what A closure would be. – fleablood Apr 08 '16 at 21:22
  • @fleablood I actually mean, $A$ not closed, because $\mathbb{N}$ and $\mathbb{R}$ are open anyway. – ahorn Apr 08 '16 at 21:23
  • @ahorn There is nothing stronger: show that $\overline{f(\overline A)}=\overline{f(A)}$. – Yai0Phah Apr 08 '16 at 21:54

3 Answers3

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For instance, if $f(x) = e^x$, then $(0, \infty) = f(\overline {\Bbb R}) \subsetneq \overline{f(\Bbb R)} = [0, \infty)$.

If a function is closed, in the sense that it maps closed sets to closed ones, then it satisfies the requirement:

$$\overline A \supset A \implies f(\overline A) \supset f(A) \implies f(\overline A) = \overline {f(\overline A)} \supset \overline {f(A)}$$

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For a counter example you have to have a limit point $y\in\overline f(A)$ that is not in $f(A)$ and $y \ne f(x)$ for any limit point of $A$. As $f$ is continuous if $a_n \rightarrow x$ then $f(a_n) \rightarrow f(x)$ so for the only way for this to happen that I could see we'd need $a_n$ diverge but $f(a_n)$ converge.

So my thought was to have a (partially) unbounded $A$ that doesn't have boundery points wheres $f(A)$ is bounded and does.

My thought was to manipulate $A = \mathbb N$ (because {n} diverges as $n \rightarrow \infty$) and $f(x) = 1/x$ (because {1/n} converges as $n \rightarrow \infty$) so $0 = \lim_{n\rightarrow \infty} 1/n \in \overline {f(A)}$ but $0 \not \in f(\overline A)$ .

But I like the other posters' answers of $f(x) = e^x$ better which is the same idea. Namely $\mathbb R$ has no lower bound but $f(\mathbb R)$ does.

(I think I'd need some tweaking as $f(x) = 1/x$ isn't continuous and f(0) isn't defined if I include 0 in the domain. I guess this isn't a deal breaker but it didn't feel right to not have 0 in the domain but to have it in the range space.

(I finally used $f(x) = x$ if $x \le 1$ and $2- 1/x$ if $x > 1$ so $f: \mathbb R \rightarrow (-\infty, 2) \subset \mathbb R$ so for $A = \mathbb N$ $f(A) = f(\overline A) = \{2 - 1/n\}$ whereas $\overline {f(A)} = \{2 - 1/n\}\cup \{2\}$.)

(But I think $f(x) = e^x$ is cleaner and neater.)

fleablood
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  • I'd prefer it if you put your answer in the context of a (general) metric space - boundedness only makes sense with a real space. But, nevertheless, it makes sense. – ahorn Apr 08 '16 at 21:37
  • Bounded in non real metric spaces means there is an M so that all d(x,y) < M. That'd hold in any metric space. Let A be unbounded set in X but f(A) bounded in a complete Y. should work. – fleablood Apr 08 '16 at 21:49
  • Okay, I take back my comment. So M could be any real number $\geq$ the diameter of the set? – ahorn Apr 08 '16 at 21:56
  • I think so. And unbounded set would have not diameter. I think limit points map to limit points so you'd need something where the image has boundary points "where" the pre image doesn't. I could be wrong but I can only imagine this as where the preimage is unbounded but the image is. Hence why they call it a boundary. Or in other word y_n-> y but f_inv(y_n)-> infinity. Once you think in those terms counter examples are more easily conceivable. – fleablood Apr 08 '16 at 22:06
  • @fleablood No. Show that, for all metric space $(X,d)$, there is a metric $d'$ on $X$ such that $d,d'$ induces the same topology (the same collection of open sets) and $d'$ is bounded, namely $\exists M,\forall x,y\in X\colon d'(x,y)\le M$. Hint: Consider $d'=min(d,1)$. – Yai0Phah Apr 08 '16 at 22:15
  • @FrankScience No what? – fleablood Apr 08 '16 at 22:33
  • @FrankScience I prefer d' = d/(d+1) but why am I supposed to be doing this? – fleablood Apr 08 '16 at 22:36
  • @fleablood On what you said about the unboundedness of the preimage. The boundedness is meaningless. – Yai0Phah Apr 09 '16 at 08:23
  • How so? We need y to be limit point of but not in f(A) such that f_inv(y) (if it exists) is not a limit point of A. I don't we can allow f_inv(y) to exist and not be a limit point of A. It's easy to imagine that f_inv(y) doesn't exist. One simple way is if A is unbounded while f(A) is. It isn't a guarantee. As you point out id:R,d -> R,d' maps unbounded R to bounded R which still has no boundary points unless we "put" R,d' "inside" something else. Example if d' = d/(d+1) for reals and d'($\pm\infty$,x)=1 then id:R,d->extended R works. as cl(F(R)) = ext R $\ne$ f(cl(R)) = f(R) = R. – fleablood Apr 09 '16 at 17:11
  • boundedness is meaningless without a metric. Sure, who the @@@@ ever said otherwise? – fleablood Apr 09 '16 at 17:13
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The continuous image of a closed set is not necessarily closed.

Example 1: Let $X=R^2$ and $Y=R$ with the usual topologies. The projection $f:(x,y)\to x$ to the first co-ordinate is continuous. The set $\{(x,1/x): x>0\}$ is closed in $R^2$ but its image is $(0,\infty)$ which is not closed in $R.$

Example 2: Let X=(0,1] and Y=[0,1] with the usual topologies. Let $f$ be the identity embedding $(f(x)=x)$ of $X$ into $Y$. Then $X$ is certainly closed in $X$, but its image is not closed in $Y.$

DanielWainfleet
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