Proving that there exists an uncountable number of distinct basis of the euclidean topology on $\mathbb R$

Question

In my general topology textbook there is the following exercise:

• (i) - Let $\mathcal B$ be a basis for a topology $\tau$ on a non-empty set $X$. If $\mathcal B_1$ is a collection of subsets of $X$ such that $\mathcal B \subseteq \mathcal B_1 \subseteq \tau$, prove that $\mathcal B_1$ is also a basis of $\tau$
• (ii) - Deduce from (i) that there exists an uncountable number of distinct basis of the euclidean topology on $\mathbb R$

I already proved statement number (i), but I'm having some trouble to think of a proof for the second statement.

I think that if we consider $\mathcal B= \{]a,b[:a,b\in \mathbb R\}$ as a basis for the euclidean topology $\tau$, then we just need to show that there exists an uncountable number of sets $\mathcal B_1$ such that $\mathcal B \subseteq \mathcal B_1 \subseteq \tau$. But how can I show that?

Answer 1

You have the right idea. Suppose we can find an uncountable collection of open sets not contained in the standard basis for the euclidean topology. Denoting this collection by $\mathcal U$, we would then have that for any $U \in \mathcal U$, $\mathcal B \subset \mathcal B \cup U \subset \tau$, which would give us the result we want.

In comparison with $\tau$, $\mathcal B$ is rather small - the vast majority of open sets in the euclidean topology are not open intervals. This is why theorems which let us work with basis elements rather than open sets are so useful - they greatly decrease the number and types of sets that we have to consider.

Finding a collection $\mathcal U$ which works is therefore not so difficult. For example, all sets of the form $(a, \infty),\, a \in \mathbb R$ are open, but none of them appear in the standard basis.

Answer 2

You are on the right track. So you have

$$\mathcal B= \{(a,b):a,b\in \mathbb R\}$$

Now for any $x\in\mathbb{R}$ add a single open set to $\mathcal B$:

$$\mathcal B_x= \mathcal B\ \cup\{(x,\infty)\}$$

What can you say about $\mathcal B_x$? Is it a basis? How many different $\mathcal B_x$ are there?

Answer 3

One base for $\Bbb R$ is the countable family $\mathcal{B}=\{(a,b): a < b, a,b \in \Bbb Q\}$.

Note that for any irrational $w$, $\mathcal{B}(w):=\mathcal{B} \cup \{(0,w)\}$ is a family of open sets as described in (i). So for every irratonal $w$, we have a different new base $\mathcal{B}(w)$ and as the are uncountably many irrationals, we have at least that many bases for $\Bbb R$.