connected and compact but not closed?

Question

Take $A= f(B)$ contained in a topological space $X$, where $B= \{(x,y)\in {\Bbb R}^2| 2\leq x^2+ y^2\leq 6\}$ and $f:{\Bbb R}^2 \to X$ is a continous map. How this one is not closed but is compact? It is compact and connected. But when it is compact why not it is closed then?

Answer 1

B is a compact, closed, connected annulus. By continuity f(B)=A is compact and connected. However to guarantee f(B) is closed we would need the function f to map closed sets to closed sets, which is not a consequence of continuity. Consider X to be the indiscrete topology on R^2, then any map is continuous so you can take f to be the identity map and B is not closed in the indiscrete topology.

Answer 2

Consider the case where $X=\{1,2\}$ where $\{1\}$ is open but $\{2\}$ is not open. (This is called Sierpinski space.) Suppose $A= f(B)=\{1\}$. Then $A$ is not closed in $X.$