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Which properties hold for the following set?

Open, connected, compact, closed.

$A=f(B)\subset X$ where $B=\{(x,y) \in\mathbb{R}^2: 1\leq x^2+y^2\leq 2\}$,
$X$ is an arbitrary topological space and $f: \mathbb{R}^2\to X$ is an arbitrary continuous map.

My thoughts:
The given set $B$ is closed annulus and it is connected and also compact. Since $f$ is continuous so $A$ will compact and connected.
But how can I verify that closed /open/both/neither.

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up vote 1 down vote accepted

A continuous map does not have to be neither open nor closed. Here by an open map I mean the one which takes open sets to open sets, and the closed map - the one which takes closed sets to closed sets.

As for counterexamples, let $X = \{a,b\}$ with the trivial topology $\{\emptyset,X\}$ and define $f:B\to X$ to be a constant $b$ map. Clearly, $f(B) = b$ which is neither open nor closed.

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However, if $X$ is Hausdorff, $A$ will be a closed set by compactness. – Olivier Bégassat Jan 2 '13 at 12:52
Please fix ambiguous terminology: to say a map is open or closed usually refers to its graph having such properties. – akkkk Jan 2 '13 at 12:53
@akkkk: done, is it better now? – Ilya Jan 2 '13 at 13:18

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