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 $?$

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

Answer 1

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 $(0, 1)$.)

Answer 2

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.