Closedness of the subset [0,1] in $\Bbb{Q}$

Question

I'm working through the definitions given in 2.18 in Baby Rudin, and I was wondering how to prove that $E = [0,1] \cap \Bbb{Q}$ is closed in $\Bbb{Q}$.

Clearly, the set $[0,1] \cap \Bbb{R}$ is closed in $\Bbb{R}$, since for all neighborhoods of a point $p \in [0,1]$, we can always find a $q \neq p$ such that $q \in [0,1]$. Intuitively, this is because for all $p \in [0,1] \cap \Bbb{R}$, we can always find a $q \in [0,1] \cap \Bbb{R}$ such that $p - \epsilon < q < p$ or $p < q < p+\epsilon$ for all $\epsilon >0$.

However, that the set $E = [0,1] \cap \Bbb{Q}$ is closed in $\Bbb{Q}$ is sort of clear, but not intuitive to me, since $\Bbb{Q}$ is not uncountably infinite as $\Bbb{R}$ is. Does one have to use that fact that it is closed in $\Bbb{R}$ and that $\Bbb{Q}$ is dense in $\Bbb{R}$? How would one solve it from first principles? i.e. Something along the lines of: Consider an arbitrary $p \in [0,1] \cap \Bbb{Q}$, we show that it is a limit point by...

Any help would be greatly appreciated!

Answer 1

I'm sorry to say I couldn't follow your argument that it is intuitive $[0,1]$ is closed in $\mathbb R$ at all.

$[0,1]$ is closed (in $\mathbb R$) because all its limit points are in in $[0,1]$. That is; If $p \in \mathbb R$ and for every $\epsilon > 0$ it will always be that the interval $(p-\epsilon, p+\epsilon)$ will contain a point $q \in [0,1]$, then we can conclude that $p$, itself, is in $[0,1]$.

Pf: If $p < 0$ and $\epsilon = 0 - p= |p|$ then $(p-\epsilon, p + \epsilon) = (2p, 0)$ is disjoint from $[0,1]$ and there is no point, $q$, in common with $[0,1]$. And if $p > 0$ and $\epsilon = p - 1$ then $(p-\epsilon, p+ \epsilon) = (1, 2p - 1)$ and there is no point, $q$, in common with $[0,1]$.

So if $p$ is a limit point (if there are any limit points, there might not be) then it would have to be that $0 \le p \le 1$ so $p \in [0,1]$.

Thus by definition $[0,1]$ is closed.

.....

But notice there is nothing in that proof that pertains to any property of $\mathbb R$ and it would have held true for any subspace of $\mathbb R$ at all. Everything we said would be true for $\mathbb Q$ or any $B \subset \mathbb R$.

Also, although the proof to show $[0,1]$ was closed was particular to $[0,1]$, there is no proof of any other set being closed that would rely upon the space.

Claim: If $A$ is closed in space $X$ then $A\cap Y$ is closed in $Y \subset X$.

Pf: Let $p \in Y$ be a limit point of $A\cap Y$ then every neighborhood, $N$ of $p$ contains a point $q; q \ne p; q \in A\cap Y\subset A$. But then $N$ is a subset of a neighborhood $N'\subset X$. $q \in A$ and $q \in N'$. So $p$ is a limit point of $A$. And as $A$ is closed in $X$ that means $p \in A$. And as $p \in Y$. So $p \in A\cup Y$. So $A\cup Y$ is closed in $Y$.

Answer 2

The density of $\mathbb Q$ plays no role here. If $A$ and $B$ are subsets of $\mathbb R$ and $A$ is a closed subset of $\mathbb R$, then $A\cap B$ is a closed subset of $B$. That's so because the complement of $A\cap B$ in $B$ is equal to $A^\complement\cap B$. Now, since $A$ is a closed subset of $\mathbb R$, $A^\complement$ is an open subset of $\mathbb R$ and therefore $A^\complement\cap B$ is an open subset of $B$.

Answer 3

A bit formal :

Baby Rudin: Theorem 2.30.

Suppose $Y \subset X$. $A$ subset $E$ of $Y$ is open relative to $Y$

$\iff$

$E=Y \cap G$ for some open subset $G$ of $X$.

$B= [0,1]$ is closed in $\mathbb{R}$.

$B^c:= \mathbb{R}$\ $[0,1]$ is open in $\mathbb{R}$.

Hence

$(B^c \cap \mathbb{Q})$ is open in $\mathbb{Q}$.

The complement of $(B^c \cap \mathbb{Q})$ in $\mathbb{Q}$ is closed in $\mathbb{Q}$ :

$\mathbb{Q} - (B^c \cap \mathbb{Q})=$

$\mathbb{Q} \cap (B^c \cap \mathbb{Q})^c =$

$\mathbb{Q} \cap (B \cup \mathbb{Q}^c)=$

$(\mathbb{Q} \cap B)\cup (\mathbb{Q} \cap \mathbb{Q}^c)=$

$\mathbb{Q} \cap B$.