Show that $A:=\left \{ \left(t,\frac{1}{t}\right) : \frac{1}{4} 
$\mathbb{R}^2$ is given with $|\cdot |_\infty$. Is $A:=\left \{
\left(t,\frac{1}{t}\right) : \frac{1}{4} <t <4\right \} $ open or
  closed?
I suppose it is not open. I visualized the set the following:

I know that there can be no $\varepsilon$-Ball such that $B_\varepsilon(x)\subset A$ for all $x\in A$. I don't really know how to justify/prove it though...
Any suggestions?
 A: It is not  open since you then cannot be an open square inside the curve.
Indeed take the point $p=(1,1)\in A$ and assume that there exists an open square  $B$ centered at $p$ with  diameter $r$.
Take the point $q=(1+\epsilon,1+\epsilon)$ for some $\epsilon>0$ small enough such that  $q \in B$.
Note that  $q$ has not equal to a point   $(t_0,\frac{1}{t_0})$ for some $t_0 \in (\frac{1}{4},4)$ because then we would have that $(1+\epsilon)=\frac{1}{1+\epsilon}$ thus $\epsilon=-2<0$ which is a contradiction.
Also it is not closed.
Take the sequence $x_n=(t_n,\frac{1}{t_n})$ for $t_n=\frac{1}{4}+\frac{1}{n}$ which converges to $ (\frac{1}{4},4) \notin A$
with respect to $|\cdot|_2$.
But $\forall x \in \Bbb{R}^2$ we have that $|x|_{\infty} \leq|x|_2$
A: It is not open because $(1,1)\in A$ but, for any $r>0$, the open ball $B_r\bigl((1,1)\bigr)$ is not a subset of $A$ (for instance, $\left(1,1+\frac r2\right)\in B_r\bigl((1,1)\bigr)\setminus A$).
And it is not closed because, since it is bounded, if it was closed it would be compact. But it is clear that $A$ is homeomorphic to $\left(\frac14,4\right)$ (just consider the map from $A$ onto $\left(\frac14,4\right)$ defined by $(x,y)\mapsto x$), which is not compact.
