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!

  • $\begingroup$ 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

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 $x \in V \subseteq U$?

  • 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

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