To check whether the set $S$ is compact.

Question

The set $S=\{(x, y):x=0,|y|<=1\}\cup \{(x, y):0<x \le 1,y=sin(\frac{1}{x})\}$.We need to check whether $S$ is compact.

1) $S_1 = \{(x, y):x=0,|y|<=1\} $.Now, $S_1$ is closed and bounded subset of $\mathbb{R}^2$ ,hence it is compact.

2)$S_2= \{(x, y):0<x \le 1,y=sin(\frac{1}{x})\}$.

Now,I know this theorem that

if $E$ is a compact set then $f$ is continuous on $E$ iff the graph of $f$ is compact.

We know that value of $x$ comes from the interval $[0,1]$

However we don't have the function $sin(\frac{1}{x})$ continuous in the interval [0,1] even if we define $sin(\frac{1}{x})=0$.So if graph ($S_2 \cup {(0,0)}$)becomes compact then sin($\frac{1}{x}$) becomes continuous on the interval $[0,1]$,which is a contradiction.

Hence my intuition is that $S$ is not compact. Is this OK?

Answer 1

The set $S_2$ is also compact. Note that$$S_2=\overline{\left\{\left(x,\sin\left(\frac1x\right)\right)\,\middle|\,x\in(0,1]\right\}}.$$So, $S_2$ is closed. Since it is also bounded (it is a subset of $[0,1]\times[-1,1]$), it is compact.