Question:
given,
$S= \{(x,y)∈\mathbb{R^2}: -1≤x≤1, -1≤y≤1\}$ and,
$T=S-\{(0,0)\}$
and let $f$ be continuous function from $T$ to $\mathbb{R}$, then choose the correct,
(a) Image of $f$ must be connected.
(b)Image of $f$ must be compact.
(C) any such continuous function $f$ can be extended to a continuous function from $S$ to $\mathbb{R}$
(d) if $f$ can be extended to continuous function from $S$ to $\mathbb{R}$ then image of $f$ must be bounded!
My attempt: I had done with (a) and (d) and they are correct! But I am not able to discard (b) and (c). For (b) I know, continuous image of compact set is compact, but i am not able to discard/ prove (b).
Please help me :-(, in (b) and (c) only...
I think both $(b)$ and $(c)$ are not true. My counter examples are as follows:
For $(b)$ consider the supremum norm induced from $\mathbb{R}^2$ as $f$. i.e $f(x,y):=\sup\{\vert x\vert\ , \vert y\vert \}$. Then the image of $f$ is $(0,1]$, which is not closed and not compact.
For $(c)$ consider the map $f(x,y)= \frac{1}{\sqrt{x^2+y^2}}$. This is a continuous function which is not bounded. Therefore by $(d)$ it cannot be extended to $S$.