# Which of the following condition implies that the set $A$ is compact

Question :

Let $A$ be a subset of $\mathbb R$. Which of the following properties implies that $A$ is compact $?$

1. Every continous function $f :A \rightarrow \mathbb R$ is bounded.

2. Every sequence $\{ x_n \}$ in $A$ has a convergent subsequence converging to a point in $A$.

3. There exist a continuous function from $A$ onto $[0,1]$.

4. There is no one-one and continuous function from $A$ onto $(0,1)$

In $(2)$ option we have result if $(X , d)$, is a metric space, then $X$ is compact iff every sequence has a convergent subsequence. I have no idea in other option. Please give me hint how to verify other option. Thank you

Hint:

2) is called sequential compactness (and there is a well-known theorem relating sequential compactness to compactness).

3) $\frac{1}{2}\sin(x)+\frac{1}{2}$ maps $\mathbb{R}$ to $[0, 1]$ and is continuous.

Edit

This is a hint assuming the question said: there is a continous 1-1 map from $A$ onto $(0,1)$...

4) $\arctan(x)$ is a continuous function that maps $\mathbb{R}$ to $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Better Hint

This hint was provided by @Struggler (see comments). If $A=\mathbb{N}$ then $A$ is certainly not compact and there is no continous 1-1 function from $A$ onto $(0, 1)$. (Note: if we remove the onto condition then $f(x)=\frac{1}{x+1}$ is a 1-1 map from $\mathbb{N}$ into $(0, 1)$.)

• @ Travis J : In 4), I am not understanding, what do you want to say – Struggler Jun 3 '15 at 16:09
• @Struggler, try shifting/stretching $\arctan(x)$ so that it maps $\mathbb{R}$ to $(0, 1)$. This is sort of akin to what I said for (3) shifting and scaling $\sin(x)$ so that instead of mapping to $[-1, 1]$ it mapped to $[0, 1]$. – TravisJ Jun 3 '15 at 18:45
• @ Travis : In 4), our aim to find a subset $A$ of $\mathbb R$ which is not compact in $\mathbb R$ and there is no continous one- one and onto mapping. I think set of natural number $\mathbb N$ which is not compact in $\mathbb R$ and there is no continous one -one and onto mapping from $\mathbb N$ to $(0,1)$. I think this is the counter example. – Struggler Jun 4 '15 at 11:18
• @Struggler, Ahh, I see that I did not read 4) properly. I thought it said: there was a continuous one-to-one map from $A$ to $(0,1)$. My suggestion for 4) is certainly not good then. Your example certainly satisfies no continuous 1-1 map to $(0,1)$ and it is not compact. So that seems good. – TravisJ Jun 5 '15 at 14:00
• @JessePFrancis, $\sin(x)$ maps the reals to $[-1,1]$. – TravisJ Dec 4 '15 at 0:35