2
$\begingroup$

Let $q_1, q_2, ...,$ be the rational numbers enumerated. Consider the countable collection $$\mathcal{B} = \{ B_{\frac{1}{n}}(q_i) \ | \ i,n \in \mathbb{N} \}$$ of open balls centered at rational numbers. Show that $\mathcal{B}$ is a basis for the Euclidean topology on $\mathbb{R}$.


First we need to show that $\mathcal{B}$ is a basis of a topology, so we need to check the two conditions:

(i): $x \in \mathbb{R}$ belongs to some $B \in \mathcal{B}$.

(ii): For $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2$.

For (i): This is true because there exists $q \in \mathbb{Q}$ such that $B_{\frac{1}{n}}(q) \ni x$. i.e., $(q-x) < \frac{1}{n}$.

For (ii): Suppose $B_1$ has a radius of $\frac{1}{n}$ centered at $q_1$, and $B_2$ has a radius of $\frac{1}{m}$ centered at $q_2$. Then, $|(q_1 - q_2)| < \frac{1}{n} + \frac{1}{m}$ We can find $B_3$ with radius $\frac{1}{k}$ with $\frac{1}{k} < min \{ |x-q_1|, |x-q_2|, \}$, this $B_3$ will be in $B_1 \cap B_2$ and will include $x$.

Thus, $\mathcal{B}$ is a basis that constructs a topology, say $\mathcal{T}'$.


Now I'm stuck at how to prove that this topology is the standard topology on $\mathbb{R}$, say $\mathcal{T}$.

If I want to show that the topology induced by $\mathcal{B}$, $\mathcal{T}'$ is equal to the standard topology on $\mathbb{R}$, $\mathcal{T}$.

We need to show that $\mathcal{B} \subset \mathcal{T} \implies \mathcal{T}' \subset \mathcal{T}$, and also $\mathcal{T} \subset \mathcal{T}'$. This is the part I don't know how to. Any help will be greatly appreciated.

Thanks in advance!

$\endgroup$
  • $\begingroup$ What's your definition of the standard topology on $\mathbb R$? You could just take all open intervals as the basis of $\mathcal T$ then it's easy to see they are equal. $\endgroup$ – user2345215 Mar 3 '14 at 23:45
  • $\begingroup$ Our standard topology on $\mathbb{R}$ is generated by $\mathcal{B} = (a,b) = \{ x \mid a < x < b \}$. How are they equal? @user2345215 $\endgroup$ – PandaMan Mar 3 '14 at 23:54
  • $\begingroup$ Then it's easy. Just note that 2 bases generate the same topology iff for every $x\in A\in\mathcal B_1$ there's $B\in\mathcal B_2$ such than $x\in B\subseteq A$ and vice versa. $\endgroup$ – user2345215 Mar 4 '14 at 0:01
1
$\begingroup$

Here's a hint. Let $U$ be an open set on $\mathcal{T}$. Let $x$ be any element of $U$. Is it true that we can find some set $V$ which is open in the topology of $\mathcal{T}'$ where $x \in V \subseteq U$?

$\endgroup$
  • 1
    $\begingroup$ Elements of topologies are sets, so $x \in V$ is an incorrect / absurd statement. $\endgroup$ – Xenidia Oct 24 '18 at 11:23
  • $\begingroup$ @Xenidia you are correct. Edited to fix this. $\endgroup$ – Bill Trok Oct 28 '18 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.