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

Question

Question :

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

Every continous function $f :A \rightarrow \mathbb R $ is bounded.
Every sequence $ \{ x_n \}$ in $A$ has a convergent subsequence converging to a point in $A$.
There exist a continuous function from $A$ onto $[0,1]$.
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

Maybe it's wor dmentioning that topological spaces with property (1) are called pseudocompact ands pace with property (2) are called sequentially compact. As you can read in the Wikipedia articles, for metric spaces these properties are equivalent to compactness. — Martin Sleziak, Jun 03 '15 at 10:29
Some related older posts on this site: If every real-valued continuous function is bounded on X (metric space), then X is compact. $K\subseteq \mathbb{R}^n$ is a compact space iff every continuous function in $K$ is bounded. Especially Mariano's answer to the latter question seems very concise and nice. — Martin Sleziak, Jun 03 '15 at 10:32

TravisJ · Accepted Answer · 2015-06-05T14:05:43.677

2

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 $(0, 1)$.)

edited Jun 05 '15 at 14:05

answered Jun 03 '15 at 10:18

TravisJ

7,426

@ Travis J : In 4), I am not understanding, what do you want to say – Struggler Jun 03 '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 03 '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 04 '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 05 '15 at 14:00
@TravisJ, why $\frac12\sin(x)+\frac12$? just $\sin(x)$ works, right? – Jesse P Francis Dec 02 '15 at 13:37
1

@JessePFrancis, $\sin(x)$ maps the reals to $[-1,1]$. – TravisJ Dec 04 '15 at 00:35

score -1 · Answer 2 · answered May 14 '22 at 06:50

-1

SIR MY DOUBT IN 1ST OPT HOW CAN IT BE CORRECT. f(x)=x ; x€(0,1) Then f(x)=(0,1) .Here f(x) is bounded in R and also continuous in (0,1)¢R. But domain is not compact.

answered May 14 '22 at 06:50

Gulab Ghosh

1

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – RiverX15 May 14 '22 at 07:10

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

2 Answers2