# Consider $A$ and $B$ of $\mathbb{R}^2$ $A =\{(x,x\sin\frac1x):x \in(0,1]\}$ and $B$=$A \cup \{(0,0)\}.$ Then which of the followings are true?

Consider the subsets A and B of $\mathbb{R}^2$ defined by

$A = \{\left(x,x\sin\frac1x\right):x \in(0,1]\}$ and $B=A \cup \{(0,0)\}.$

Then which of the followings are true?
1. $A$ is compact
2. $A$ is connected
3. $B$ is compact
4. $B$ is connected.

I know that A is connected but not path connected, so 2 is true. But I'm not sure about the others.

-
For 1., note that $(0,0)$ is in the closure of $A$, but not in $A$. For 3., note $B$ is closed and bounded. If you believe that $A$ is connected, you should be able to see that $B$ is also (since $B=\overline A$). (Why is $A$ not path connected?) – David Mitra Dec 21 '12 at 3:07

I claim that $(2)-(4)$ are all true. Indeed, you saw that $(2)$ is true, but consider the function $f:[0,1]\to B$ defined by $$f(x)=\begin{cases}x\sin\left(\frac{1}{x}\right) &:\: x\in(0,1]\\ 0 &:\: x=0\end{cases}.$$ It is routine to show this is a continuous function, hence its image is compact and connected.

-
I've tried something different. Would you please take a look? – Sugata Adhya Dec 21 '12 at 6:01

The following proposition is is useful

Proposition: If $N$ and $M$ are connected sets which are not separated, then $N \cup M$ is connected.

Separated sets: Two subsets $N$ and $M$ of a topological space $X$ are said to be separated if

(i) $N$ and $M$ are disjoint, and

(ii) neither contains an accumulation point of the other.

-
I've tried something different. Would you please take a look? – Sugata Adhya Dec 21 '12 at 6:01

{${1\over x}:x\in(0,1]$} $\subset[2\pi,4\pi]\implies$ {$\sin{1\over x}:x\in(0,1]$} $=[-1,1]\implies${$x\sin{1\over x}:x(0,1]$} $=[-1,0)\cup(0,1]\implies A=(0,1]\times [-1,0)\cup(0,1]\implies B=[0,1]\times[-1,1]$

So $A$ is not compact but connected whereas $B$ is both compact & connected.

-