# Subspace topology in proof that $\mathbb{Q}$ is connected?

I'm showing that the connected components of $\mathbb{Q}$ are sets containing a single point. Let $A \subset \mathbb{Q}$ have at least two points in it, say $p$ and $q$ with $p \neq q$ and assume without loss of generality that $p < q$. Then there exists some irrational number $x$ such that $p < x < q$. Now write $$A = (A \cap (-\infty,x)) \cup (A \cap (x, \infty)) =: U \cup V$$ It is true that $A \subset \mathbb{Q}$ and also $A \subset \mathbb{R}$. Also, $(-\infty, x)$ and $(x, \infty)$ are open in $\mathbb{R}$, so $U$ and $V$ are open subsets of $A$ with the subspace topology. Clearly, $U \cap V = \emptyset$ so $A$ is not connected. So for $A$ to be connected, it must be a singleton set.

Now for my question: $(-\infty, x)$ and $(x, \infty)$ are not open in $\mathbb{Q}$ so $U$ and $V$ are not open in the subspace topology when everything is considered as a subset of $\mathbb{Q}$. Is this not a problem for the proof? My definition for $A$ being connected is that it cannot be expressed as the disjoint union of two disjoint non-empty open sets. In the case of $A$ being connected/disconnected, the open sets of $A$ in question are $U$ and $V$. Does it not matter what we considered $A$ itself to be a subset of when using the subspace topology in the proof?

Many thanks for any help.

• What do you mean with " $(-\infty, x)$ and $(x, \infty)$ are not open in $\mathbb{Q}$ so..."? Those aren't even subsets of $\;\Bbb Q\;$ ... – DonAntonio Apr 3 '18 at 12:43

The topology of $\mathbb{Q}$ is the subspace topology induced from $\mathbb{R}$ and, for instance, $(x, \infty) \cap \mathbb{Q}$ is open in $\mathbb{Q}$, so everything is fine. You can write $(x, \infty) \cap A$ as $(x, \infty) \cap \mathbb{Q} \cap A$. When taking subspace topologies, you can easily check that it makes no difference if you take it in two steps $(\mathbb{R} \to \mathbb{Q}, \mathbb{Q} \to A)$ or in one step $(\mathbb{R} \to A)$. In particular, being connected is a property of the topological space $A$, regardless of what ambient space it is embedded into.

• Right, I understand now, thanks for your answer. To double check, would something like $C = [-\sqrt{2},\sqrt{2}]$ be closed in $\mathbb{Q}$? I think it would, because the complement of $C$ in $\mathbb{R}$ is open, so the complement intersected with $\mathbb{Q}$ is open in the subspace topology $\implies C \cap \mathbb{Q}$ closed in subspace topology. – mathphys Apr 3 '18 at 12:51
• @mathphys $[\sqrt{2},\sqrt{2}]\cap\mathbb{Q}$ is empty, and therefore, open and closed. If you meant $[-\sqrt{2},\sqrt{2}]\cap\mathbb{Q}$, then that one is also open and closed. – user547557 Apr 3 '18 at 12:54
• @mathphys No. It doesn't work in that direction. What you need is that some $A$ closed in $\mathbb{R}$ gives $A\cap\mathbb{Q}=(C\cap\mathbb{Q})^c$. In your case you can use $(-\infty,-\sqrt{2}]\cup[\sqrt{2},\infty)$. – user547557 Apr 3 '18 at 13:03
• @mathphys In the simplest form: $[-\sqrt{2},\sqrt{2}]\cap \mathbb{Q} = (-\sqrt{2},\sqrt{2})\cap \mathbb{Q}$. – 57Jimmy Apr 3 '18 at 13:08
• @mathphys Exactly! – 57Jimmy Apr 3 '18 at 13:11

$(-\infty, x)$ and $(x, \infty)$ are not open in $\mathbb{Q}$ so $U$ and $V$ are not open in the subspace topology when everything is considered as a subset of $\mathbb{Q}$.

This is very confusingly written. $(-\infty, x)$ is a subset of $\mathbb Q$, and the claim is that $U$, not $(-\infty, x)$, is open in $\mathbb Q$

To prove that $U$ is open in $\mathbb Q$, remember the definitions:

If $X$ is a topological space and $Y\subseteq X$ is equipped with the subspace topology, then a set $A\subseteq Y$ is open (in $Y$) if and only if there exists some $A'\subseteq X$ such that $A'$ is open in $X$ and $A=Y\cap A'$.

Using that definition, it is clear that for the set $U=(-\infty, x)\cap \mathbb Q$ is an open set in $\mathbb Q$, because for $U$, there exists $U'=(-\infty, x)$ which is an open set in $\mathbb R$ and $U'\cap \mathbb Q=U$.