Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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