Why the argument given by bof is not working for usual topology on $\mathbb R$?

Question

$\mathbb{R}$ with the lower limit topology is not second-countable Why the argument given by bof is not working for usual topology on $\mathbb R$?Can't we use infimum instead of minimum there?

Answer 1

No, you cannot. First of all, not every non-empty subset of $\mathbb R$ has an infimum. Besides, if $A$ is an open subset of $\mathbb R$ (with respect to the usual topology) and if $x\in A$, then $A\supset(x-\varepsilon,x+\varepsilon)$, for some $\varepsilon>0$ and therefore $\inf A$ (if it exists) is never $x$. So, there is no reason why the map $x\mapsto B_x$ from that proof should be one-to-one.

Answer 2

You're thinking about the proof in the wrong way. The argument goes as follows: suppose that $\mathcal{B}$ is any base for the lower limit (a.k.a. Sorgenfrey) topology on $\mathbb{R}$.

Then note that for each $x \in \mathbb{R}$ we have that $[x,x+1)$ is open in this topology and contains $x$ (this would already fail for the usual topology, where this set is not open and $x$ is not even an interior point of it!) so that there must be some base element $B_x$ such that $x \in B_x \subseteq [x,x+1)$ by the property fo being a base.

The crucial fact is that $x \neq y$ implies $B_x \neq B_y$. Because this implies that we have at least as many base elements in $\mathcal{B}$ as there are points of $\mathbb{R}$. And as this holds for all bases, and the reals are uncountable, there can be no countable base.

To see the crucial fact: $x \neq y$, assume e.g. $B_x = B_y$. Suppose $x < y$ first, then $x \in B_x \subseteq [x,x+1)$ and $y \in B_y \subseteq [y, y+1)$ and as $x < y$, $x \notin [y,y+1)$ and so $x \notin B_y$. But $y \in B_y = B_x$, contradiction. So $B_x = B_y$ is false. Reverse the roles of $x$ and $y$ if $y < x$. So we always have this contradiction when $x \neq y$ and $B_x= B_y$.

bof's argument shortcuts this final argument by stating that $\min(B_x) = x$, so that $B_x = B_y $ immediately implies $x = \min(B_x) = \min(B_y) = y$, so that the $\min$ function is an explicit 1-1 map from a subset of $\mathcal{B}$ onto $\mathbb{R}$ which is enough.