Choose the coorect option ?.. let S = $\{(x,y) \in R^2\mid -1 \le x \le 1 \text{ and } -1 \le y\le 1\,\}$. Let $T = S \setminus \{(0,0)\}$  be the set obtained by removing  the origin from $S$. Let $f$  be a continuous function from $T$ to  $\mathbb{R}$.
choose all the correct options:


*

*The image  of  $f$ must  be connected .

*The image  of  $f$ must be  compact.

*Any such continuous function $f$ can be extended to  a continuous function from $S$ to $\mathbb{R}$.

*If $f$ can be  extended to a continuous function from $S$ to $\mathbb{R}$ then the image of $f$ is bounded.
My attempts :  For  option a) $\mathbb{R}^2 \setminus \{(0,0)\}$  is  connected  and $\mathbb{R}$ is connected as a continuous image of connected set is  connected..so it  is true.
For option b) $\mathbb{R}$  is not compact so it is not true, as the  continuous  image of compact is compact.
option C)  is  false  because  $f(x)= \frac{1}{x}$  is not continuous at $0$
  so it is false
For  option D) is  True  because  continious  image  of  bounded set is  bounded 
Is   my reasoning  is correct /or not corrects??
Pliz tell me   or give any hints/solution
thanks in advance  

a) $T$ is indeed connected, and so $f[T]$ is connected.
b) needs a counterexample; try $f(x,y) = \frac{1}{\|x\|}$, then $f[T] = [\frac{1}{\sqrt{2}}, \infty)$ which is not compact.
c) same counterexample; $(\frac{1}{n},0)\to (0,0)$ but $f(\frac{1}{n},0) = \sqrt{n}$ has no limit.
d) is correct; if $g$ were the extension of $f$ to $S$, $f[T] = g[T] \subseteq g[S]$ where the latter set is compact, being the image of a compact space $S$. So $f$ is bounded.
