$A=f(B) \subset X$ where $B=\{(x,y)\in \mathbb{R}^2 \mid 1\le x^2+y^2\le2\},$ $X$ is an arbitrary topological space and $f:\mathbb{R}^2\to X$ is an arbitrary continuous map.Then which are true ??
A) open , B) Closed, C) Connected, D) Compact
I See that option C) is true as it is pathconnected.But how i check others option??
I assume that questions are with regard to the set $A$.
A+B) are false. Take $X=\mathbb{R}^2$ and $f(x)=x$ but now put the anti-discrete topology on $X$, i.e. only $\emptyset$ and $X$ are open/closed in $X$. The map is continuous but $f(B)=B$ is neither open nor closed.
C) holds: you are right, the image of a connected set is connected. And $B$ is connected, even path connected.
D) holds: the image of a compact space is compact and $B$ is compact as a closed and bounded subset of $\mathbb{R}^2$.
Note that if $X$ is assumed to be Hausdorff then B) holds: the compact subset of a Hausdorff space is closed and $f(B)$ is compact due to D).
In $\mathbb{R}^2$ the set $B$, which is closed annulus, is compact and connected. The continuous image of compact sets are compact. The continuous image of connected sets are connected. Hence C,D are true.
But a continuous map need not be either closed or open. For example, take $X$ to be an indiscrete space and $A$ a singleton. $f$ is continuous, since all functions to indiscrete codomains are continuous. But $A$ is neither closed nor open.
