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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

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You reasonong is correct, but you proof is of closedness of $B$ is not rigour enough –  Norbert Jun 6 '12 at 11:31
@Norbert Could you please explain? –  srijan Jun 6 '12 at 11:32
What about connectedness? –  Jonas Meyer Jun 6 '12 at 11:33
@JonasMeyer I have to edit sir. –  srijan Jun 6 '12 at 11:34
Continuous image of a connected space is connected, continuous image of a compact space is compact. This should help you with some parts of the exercise. –  Martin Sleziak Jun 6 '12 at 11:35
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1 Answer

up vote 5 down vote accepted

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.

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Sorry, I made an edit then rolled it back, because I was misreading. –  Jonas Meyer Jun 6 '12 at 11:42
@Brian Thank you very much. Intuitively it was clear to me but i was not able to prove it. Your answer is really helpful to me. Will try applying them for other problems too. –  srijan Jun 6 '12 at 11:46
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