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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:

  1. The image of $f$ must be connected .

  2. The image of $f$ must be compact.

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

  4. 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

Henno Brandsma
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jasmine
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    To disprove (b), you need a counterexample. Actually, $f(x,y)=\frac{1}{x}$ will work here too. – Dzoooks Jun 06 '18 at 00:00
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    Your reasons for B and D are fantasy. – William Elliot Jun 06 '18 at 02:02
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    For (b) your error in your reasoning is that although the continuous image of a compact space is compact, the continuous image of a non-compact space $T$ is not necessarily non-compact. For example if $A= {1}$ is a one-point space then there is just one function $ f:T\to A$; it is continuous and its image $A$ is compact . – DanielWainfleet Jun 06 '18 at 05:00

1 Answers1

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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.

Henno Brandsma
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