Prove that the collection is a basis that generates standard topology of the real line

Question

Attempt:

We can use a Lemma in munkres to show this. That is, if for any open set (in usual topology) we can always find some member of collection $\mathscr{C}$ inside the open set. In other words,

Let $(c,d)$ be any open set in usual topology in the line and let $ x \in (c,d)$. If we can find some $(a,b) \in \mathscr{B}$ with $ x \in (a,b) \subset (c,d)$ this we are done.

I think if we put $a = c + \dfrac{|x-c|}{2} $ and $b = d - \dfrac{|d-x|}{2}$ then we have

$$ x \in (a,b) \subset (c,d) $$

and $a,b \in \mathbb{Q}$. Is this correct?

Answer 1

That doesn’t ensure that $a$ and $b$ are rational. For instance, suppose that $c=1$, $d=2$, and $x=\sqrt2$; then $a=\frac{1+\sqrt2}2$ and $b=1+\frac{\sqrt2}2$, both of which are irrational. Pinpointing specific rational numbers in the intervals $(c,x)$ and $(x,d)$ is a bit difficult; a much better idea is simply to use the fact that the rationals are dense in the reals, meaning that between any two real numbers there is a rational number.

That said, you also need to show that it suffices to consider open set of the form $(c,d)$. I don’t recall whether Munkres has already proved this at that point; if not, you need to start with an arbitrary open set $U$ and an arbitrary $x\in U$ and show that there are rational numbers $p$ and $q$ such that $x\in(p,q)\subseteq U$. Use the fact that the set of open intervals is a base for the usual topology on $\Bbb R$.

In view of the questions below, I’m going to add a bit of general discussion. Suppose that you have a space $\langle X,\tau\rangle$. You should think of a base for $\tau$ as a set of building blocks for $\tau$: every member of $\tau$ — i.e., every open set in $X$ — is a union of basic open sets. Suppose that $\mathscr{B}$ is a collection of subsets of $X$, and we want to know whether $\mathscr{B}$ is a base for $\tau$. The answer will be yes if two things are true:

• every $U\in\tau$ is a union of members of $\mathscr{B}$, and
• every union of members of $\mathscr{B}$ is in $\tau$, i.e., is open.

In symbols, $\tau=\left\{\bigcup\mathscr{V}:\mathscr{V}\subseteq\mathscr{B}\right\}$.

These requirements can be boiled down to ones that are in general easier to check. For the first one, let $U\in\tau$. If $U=\bigcup\mathscr{B}_U$ for some collection $\mathscr{B}_U$ of members of $\mathscr{B}$, what can we say about the sets in $\mathscr{B}_U$? Certainly they must all be subsets of $U$: otherwise $\bigcup\mathscr{B}_U$ would contain points of $X\setminus U$. On the other hand, each point of $U$ must belong to at least one of the sets in $\mathscr{B}_U$: otherwise there will be points of $U$ that are not in the union $\bigcup\mathscr{B}_U$. In other words for each $x\in U$ there must be a $B_x\in\mathscr{B}_U$ such that $x\in B_x$. We can combine these two facts in a single requirement: for each $x\in U$ there is a $B\in\mathscr{B}_U$ such that $x\in B_x\subseteq U$. And the first bullet point above can then be replaced by this one:

• For each $U\in\tau$ and each $x\in U$ there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U$.

The second requirement is easy to arrange. Each $B\in\mathscr{B}$ is trivially the union of members of $\mathscr{B}$, so in particular it must be open, i.e., belong to $\tau$. Conversely, if every member of $\mathscr{B}$ is open, then of course all unions of members of $\mathscr{B}$ are automatically open as well. Thus, we can replace the second bullet point with this one:

• $\mathscr{B}\subseteq\tau$, i.e., every member of $\mathscr{B}$ is an open set.

In short, when you want to check whether a family $\mathscr{B}$ is a base for a particular topology $\tau$, you need to check two things: that $\mathscr{B}\subseteq\tau$, i.e., that $\mathscr{B}$ consists entirely of open sets, and that for any open set $U$ and any point $x\in U$ there is some $B\in\mathscr{B}$ such that $x\in B\subseteq U$. In the present problem this amounts to verifying that each interval $(p,q)$ with rational endpoints is an open set in the usual topology, which is trivial, and verifying that if $U$ is any open set in the usual topology, and $x\in U$, then there are rational numbers $p$ and $q$ such that $x\in(p,q)\subseteq U$, which you did in an earlier comment.

If you tried to do this with the set of intervals $[p,q)$ such that $a,b\in\Bbb R$ and $p<q$, you’d be able to show that if $U$ is an ordinary open set, and $x\in U$, there are $a,b\in\Bbb R$ such that $x\in[a,b)\subseteq U$, but you would not be able to show that these sets $[a,b)$ are open in the usual topology, because they aren’t. You would in fact have a base for a topology on $\Bbb R$, but not the usual one: you would have a base for the lower limit topology.

Answer 2

No, this does not work. There is no reason why $a,b \in \mathbb{Q}$. Also not every open set of $\mathbb{R}$ with the usual topology is an interval $(a,b)$ (for example, consider the disjoint union of two open intervals).

A good proof will use that $\mathbb{Q}$ is dense in $\mathbb{R}$.

Answer 3

Let any point $x \in \mathbb{R}$ then there exists a basis element $(a,b)$ containing $x$. i.e., $a<x<b$ . Since if $g,h\in \mathbb{R}$ such that $g<h$ then there exists a rational number $r$ such that $g<r<h$. Using this fact we have $a<m<x<n<b$ where $m,n\in \mathbb{Q}$

Thus $x\in (m,n)\subset (a,b)$