-1
$\begingroup$

Let $A=f(B)\subset X$ where $B=\{{(x,y) \in \mathbb{R}^2 \mid 1 \leq x^2+y^2 \leq 2 \}} $, $X$ is arbitrary topological space and $f: \mathbb{R}^2 \to X$ is an arbitrary continuous map. Then $A$ is compact and connected but not closed and open.

I have a doubt that if $A$ is compact then it is closed, but the answer says only compact and connected. Please help me as I have learnt topology several years back and I am not that strong in it now.

$\endgroup$
  • $\begingroup$ To conclude closedness from compactness, the space has to be Hausdorff, but this holds for any metric spaces. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 7 '18 at 15:08
  • $\begingroup$ @GNUSupporter Note that $A\subset X$ with $X$ a topological space, so Hausdorff cannot be used here (maybe you're just stating facts to refresh OP's memory, which is nice). $\endgroup$ – The Phenotype Feb 7 '18 at 15:10
  • $\begingroup$ @GNUSupporter So compact doesn't imply closed if space is not Hausdorff? $\endgroup$ – zafran Feb 7 '18 at 15:11
  • $\begingroup$ The statement "closed and bounded$\iff$compact" is the Heine Borel Theorem for Euclidean spaces. For more general topological spaces, this need not hold true. $\endgroup$ – Prasun Biswas Feb 7 '18 at 15:14
  • 1
    $\begingroup$ To show compactness and connectedness, note that $A$ is the continuous image of a compact and connected set ($B=$ the annulus bounded by the circles with radii $1$ and $\sqrt 2$ centered at origin in $\Bbb R^2$). $\endgroup$ – Prasun Biswas Feb 7 '18 at 15:17
2
$\begingroup$

$A$ can be open and closed: take $X=\{0\}$ with any topology you like. Then we have that $f$ is constant, so continuous, and $A=f(B)=\{0\}=X$ is open and closed as it is the space itself.


Now assuming one of the equivalent statements: $f$ is injective or $f^{-1}(A)=B$:

Suppose that $A$ is closed and open, then due to the continuity of $f$ we would have that $B=f^{-1}(A)$ is closed and open, but obviously $B$ is not open is $\mathbb{R}^2$ (take for example any open ball around $(2,0)$, this will always have a non-empty intersection with $B^c$), so we have a contradiction.


Remarks:

  • OP probably already knew it, but the image of a compact set under continuous function $f$ is also compact. Exactly the same holds if you replace the words compact by the words connected.

  • Taken from one of my comments: Here is the proof of compact implies closed in Hausdorff spaces and here is a counterexample for general topological spaces.

  • Special thanks to @AndreasBlass for noting that injectivity is needed to make my proof hold.

$\endgroup$
  • $\begingroup$ But it is closed in $\mathbb{R^2}$. $\endgroup$ – zafran Feb 7 '18 at 15:18
  • $\begingroup$ @zafran Indeed, but you need to show it is not (open and closed), so not open or not closed. $\endgroup$ – The Phenotype Feb 7 '18 at 15:19
  • $\begingroup$ The question doesn't say that $f$ is one-to-one, so it need not be the case that $B=f^{-1}(A)$. $\endgroup$ – Andreas Blass Feb 7 '18 at 18:51
  • $\begingroup$ @AndreasBlass You're right, I made the fallacy that $f^{-1}(A)=\{b\in B\ |\ f(b)\in A\}$ instead of the fact that $f^{-1}(A)=\{b\in \mathbb{R}^2\ |\ f(b)\in A\}$. $\endgroup$ – The Phenotype Feb 7 '18 at 20:29
0
$\begingroup$

Compactness and connectedness are preserved by continuous maps, thus the compactness and connectedness of $B$ imply the same properties on $A$, its image.

It is not necessarily open as the identity $\mathbb{R}^2 \to \mathbb{R}^2$ shows.

In most common cases, to be honest, the image is indeed closed. The hypothesis you have to require is $X$ to be an hausdorff space: every two points can be separated by opens. Precisely, for every $x,y \in X$ there exist disjoint opens $U,V$ such that $x \in U, y \in V$. This property is shared by almost everything geometric that come to your mind.

Still, there are counterexamples to "compact $\Rightarrow $ closed" without the hausdorff property. Define $X$ in this way. Its points are real numbers plus a special point, $z$. Its closed sets are a finite number of real points ( $z$ is not allowed) or the entire $X$. Then the one-point-set $\{z\}$ is of course compact because every open covering is made of one open (it is just one point!), but it is not closed: its closure is the whole $X$! The other closed sets do not contain $z$.

This example arise often in algebraic geometry. It is like we have added the real line itself as $z$ to the set of real numbers, and we declared it near to every point (every open contains $z$). This allow algebraic geometrists to make interesting tricks :)

In hausdorff case, the proof that a compact is closed is easy. Try to figure it out geometrically.

Let $K \subset X$ be a compact set, and let $p\not \in K$. We must show that there is an open set that contains $p$ and does not intersect $K$ (i.e., the complement is open). By hp, for every $x \in K$ there exist $x \in U_x , p \in V_x$ open sets which separate $x,p$. $\{U_x \}_x$ is an open covering of $K$, thus we can extract a finite covering $U_1, .., U_n$. Then the corresponding $V_i$ have intersection $V= \bigcap V_i$ which contains $p$ and it is disjoint by $\bigcup U_i \supset K$. Yuppi!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.