A problem of compactness and connectedness

Question

Consider the subset $A$ and $B$ of $\mathbb{R}^2$ defined by $A =\{(x, x\sin\frac{1}{x}) :x\in(0,1]\}$

$B = A\cup \{(0,0)\}$

I have to check for compactness and connectedness of $A$ and $B$.

Here is my attempt.

$A$ is bounded but not closed as 0 is the limit point of set $A$ but it doesn't belongs to $A$. Hence $A$ is not compact.

$B$ is compact since it is closed and bounded subset of $\mathbb{R}^2$.

I am not able to figure out connectedness of given sets.

Am I correct? Is there any other way to tackle this problem? I need help with this.

Thank you very much

Answer 1

I think you mean that $A$ is not closed because $(0,0)$ is a limit point of $A$ that is not in $A$, not $0$. That’s correct, and $B$ is closed and therefore compact, though you haven’t really justified the assertion that it’s closed.

Note that the function $$f:(0,1]\to\Bbb R^2:x\mapsto\left(x,x\sin\frac1x\right)$$ is continuous, and $(0,1]$ is connected; what does that tell you about the connectedness of $A$?

Can you show that the function $$f:[0,1]\to\Bbb R^2:x\mapsto\begin{cases}f(x),&\text{if }x\in(0,1]\\0,&\text{if }x=0\end{cases}$$ is also continuous? If so, that gives you an easy way to see that $B$ is both compact and connected, because $[0,1]$ is both compact and connected.