Why is the following set neither open nor closed?

Question

I would have said answer $(ii)$, because I though it was between 0 and infinity but the correct answer is $(iv)$ which I don't get why.

I have really trouble with these types of exercises. Why is it countable here? Why is a countable set necessarily not open?

thanks

Answer 1

You are right, it is not closed because you can find a sequence of points of $E$ that converges to a point outside of $E$.

It is also not open, since $E$ has at most a countable number of points. Since it is a subset of $\mathbb R^2$, it can not be open (take a point of $E$, if there is a ball that is contained in $E$, it would have an uncountable number of points).

Answer 2

While cardinality provides one way of seeing that $E$ isn't open, it isn't the only way. Another fairly easy way is to note that $A = \{(x,y)| x \not \in \mathbb Q \}$ is dense in $\mathbb R ^ 2$ but is disjoint from $E$. That would be impossible if $E$ were open.

Answer 3

$E$ is not closed since it does not contain $(0,0)$ which is a limit point of $E$.

$E$ is not open, since it contains $(1,1)\in E$, but $E$ does not contain any disk (or ball) centred at $(1,1)$. In particular, for every $(x,y)\in E$, we have that $x\le 1$. If $E$ was open, then there would be points $(x,y)\in E$, with $x>1$.