Is this set $A$ open?

Question

I have the set $A =$ {$z \in \mathbb C: z = x + 0i = x, x \in \mathbb R$}. Is this set open in the complex plane?

The set $A$ contains all the points on the real axis in the complex plane. This set is open if $A$ contains all of its interior points, i.e. it contains all points safely inside the set $A$. But an open ball doesn't exist around any of the points in $A$, so surely $A$ contains no interior points, and hence $A$ is not open?

Your reasoning is almost correct. What you need to prove is that for a point, say $0$, and every $r > 0$, $ B(0, r) \not \subseteq A$, which you can prove by what you descrived in the last part of your question. — Victor, Feb 29 '24 at 17:19
@Victor Thanks, that makes sense intuitively too. Was I also right that the set of interior points of $A$ is empty? — adisnjo, Feb 29 '24 at 17:20
Yes, its interior is empty, which you can prove by using the same argument as I described for every point in $A$. — Victor, Feb 29 '24 at 17:22

score 2 · Accepted Answer · answered Feb 29 '24 at 17:21

The set $A$ is open if every point in $A$ is interior, meaning for each point in $A$, there is a (non-empty) open ball around that point which is completely contained in $A$. For the set $A$ you have, which is the real axis, a non-empty ball of radius $r$ around any point $x$ in $A$ will contain points outside of $A$, like $x+ir/2$, for example. So $A$ is not open.

score 1 · Answer 2 · answered Feb 29 '24 at 17:31

I don't know your level of knowledge, but here is a more sophisticated way to reason, maybe it's futile complication but you'll be the judge...

Consider the function $f:\mathbb{C}\to \mathbb{R}: a+ib\mapsto b$. This is clearly a continuous function, and $f^{-1}(\{0\})=A$. $\{0\}$ is closed in $\mathbb{R}$ and so must be $A$ (preimage of a closed set through a continuous function is closed).

This doesn't rule out that $A$ isn't open but $\mathbb{C}$ is connected, which means that only $\mathbb{C}$ and $\emptyset$ are the subsets open and closed, clearly $A\neq \mathbb{C},\emptyset$ so $A$ can't be open.

Is this set $A$ open?

2 Answers2