In my general topology textbook there is the following exercice:
Prove that $\mathbb R$, $\mathbb Z$, $\mathbb Q$, and $\mathbb I$ are regular topological spaces.
My proof
$\mathbb Z$, $\mathbb Q$, and $\mathbb I$ are subspaces of $\mathbb R$, so if $\mathbb R$ is regular, then those subspaces will also be regular.
So, let $A\subseteq \mathbb R$ be a closed subset, and $x \in \mathbb R \setminus A$.
We can write $A = \bigcup_i A_i$, for some index set $i \in I$, where $A_i$ are the connected components of the set $A$.
We can define $\varepsilon = \inf \ \{ |x – \delta|, \delta \in A\}$.
Then we have that $x \in (x – \frac{\varepsilon}{2} ,x + \frac{\varepsilon}{2})$, and $\forall i \in I, A_i \subseteq (\inf A_i – \frac{\varepsilon}{2}, \sup A_i + \frac{\varepsilon}{2})$, so we conclude that $A \subseteq \bigcup_i(\inf A_i – \frac{\varepsilon}{2}, \sup A_i + \frac{\varepsilon}{2})$. Is trivial that $\bigcup_i(\inf A_i – \frac{\varepsilon}{2}, \sup A_i + \frac{\varepsilon}{2}) \cap (x – \frac{\varepsilon}{2} ,x + \frac{\varepsilon}{2}) = \emptyset$, thus the set $\mathbb R$ is regular.
So, first of all I would like to know if My proof is correct and if so, is there anything that I can do to improve it?
The one thing that bothers me about this proof is that, in order to define $\varepsilon$, I assumed that $\mathbb R$ is a metric space with $d(x,\delta) := |x – \delta|$. This is a problem because at this point I have not learned about metric spaces because My textbook starts with topological spaces and then moves on to metric spaces. How can I prove this without using any metric and instead use the definitions and axioms of topological spces?
Best Answer
There is no need at all to use connected components.
Let $x \in \Bbb R$ and $A \subseteq \Bbb R$ be closed. Then $\Bbb R\setminus A$ is open and contains $x$, so there is an open interval $(a,b)$ that contains $x$ and is a subset of $\Bbb R\setminus A$ (so also $A \subseteq \Bbb R \setminus (a,b)$.)
Now let $c,d$ be reals such that $a < c < x$ and $x < d < b$, which can always be done (the reals are "order dense"). Now $(c,d)$ is an open neighbourhood of $x$ and $V:=\Bbb R\setminus [c,d] \supseteq \Bbb R\setminus (a,b) \supseteq A$, and clearly $(c,d) \cap V = \emptyset$.