In an exercice I am asked to prove the following:
Prove that the Sorgenfrey Line does not satisfies the second axiom of contability.
This is my second proof for this exercise because the first one was wrong. Because I'm studying topology by my self and the book has no solutions I'm share my proof hopping that someone could correct me if I made some mistake or if the proof is not valid.
My proof:
Let $\cal B$ be a basis for the Sorgenfrey Line.
Then we have that $\forall U \in \tau, \forall x \in U, \exists B \in \mathcal B: x \in B \subseteq U$.
So let $U_a = [a,2)$, with $a \in [0,1]$, then $U \in \tau$
So $\exists B_a \in \mathcal B: a \in B_a \subseteq U$.
Now let's define the set $S = \{B_a:a \in [0,1]\}$, Then we can define a bijective function $f: [0,1] \to S$ with $f(x) = B_x$, so we have that $S \sim [0,1]$, which means that $S$ is uncountable.
We know that if a set $A$ is countable then every subset $B$ of $A$ is also countable. We know that $S \subset \mathcal B$, and $S$ is uncountable, which means that $\mathcal B$ cannot be countable, thus proving that does not exists a countable basis for the Sorgenfrey Line.
Is this proof correct?
You have to actually show $f: [0,1]\to \mathcal{B}, a \to B_a$ is an injection (the set $S$ is unnecessary ): suppose $a \neq a'$. Then $a < a'$ WLOG. But then $a \in f(a)$ and $a \notin f(a')$ shows that $a \in f(a)\setminus f(a')$ so that $f(a) \neq f(a')$ (as sets).
And the existence of an injection from $[0,1]$ into $\mathcal{B}$ is a direct proof that $|\mathcal{B}| \ge |[0,1]|$ and so all bases of the Sorgenfrey line have size at least continuum.
BTW, minor point, we don't need to use $[0,1]$ and sets $[a,2)$ but $\Bbb R$ and taking $B_x$ with $x \in B_x \subseteq [x,x+1)$ is simpler to me and more "uniform". The same arguments apply to this choice as well.