Is it closed and compact? $A= f(B) \subseteq X$ where  $B =\{(x,y)\in \mathbb{R}^2 |1≤ x^2 +y^2 ≤ 2\}$. $X$  is an arbitrary topological space and $f :\mathbb{R}^2 \rightarrow X $ is an arbitrary continuous map.
which of the following  is correct.


*

*open

*closed

*compact

*connected
I know that  continuous image of a closed set is closed  as continuous image of compact is compact so  only  option 2 and option 3 is trues and correct.
Is it correct ? tell  me  where I'm wrong....I would be more thankful who give me hints or any solution.....thanks in advance 

$B$ is compact $\implies$ $f(B)$ is compact $(\because$ continuous image of compact set is compact.)
