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.

  • $\begingroup$ 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
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    $\begingroup$ 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

$A$ 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.


  • 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.

  • $\begingroup$ 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

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!


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