Exist a norm such that $S= \{x= (x_{1},x_{2}) \in R^{2}| \|x\| \leq 1, x_{2} >0\}$ which isn't open nor closed?

Question

Let $X = R^{2}$ and $\|.\|$ a norm in X:

Show that exist a norm such that $S= \{x= (x_{1},x_{2}) \in R^{2}| \|x\| \leq 1, x_{2} >0\}$ it is not open or closed.

How can I show this?

Any help?

Answer 1

This is somewhat of a trick question: For every norm $\|\cdot\|$ on $\Bbb R^2$, the set $$S=\{\,x=(x_1,x_2)\in\Bbb R^2\mid\|x\|\le 1,x_2>0\,\} $$ is neither open nor closed (under the topology induced by the norm).

In order to see that $S$ is not open, note that $(0,1)\ne(0,0)$ implies $\|(0,1)\|>0$, hence for $a:=(0,\frac1{\|(0,1)\|})$, we have $\|a\|=1$. For all $\epsilon>0$, the open ball $\{\,x\in\Bbb R^2\mid\|x-a\|<\epsilon\,\}$ fails to be a subset of $S$ because for $x:=(1+\frac\epsilon2)a$ we check that $\|x-a\|=\|\frac\epsilon2a\|=\frac\epsilon2<\epsilon$ and $\|x\|=1+\frac\epsilon2>1$.

In order to see that $S$ is not closed, i.e., that the complement of $S$ is not open, note that $(0,0)\notin S$, but every open ball $\{\,x\in\Bbb R^2\mid \|x\|<\epsilon\,\}$, $\epsilon>0$, intersects $S$. Indeed, we only need to look at $\frac\epsilon2a$ with $a$ as defined above.