I having some trouble following this passage from the Munkres Topology textbook (page 190, Chapter 4, section 30)
Example 2: In the uniform topology, $\mathbb{R}^\omega$ satisfies the first countability axiom (being metrizable). However, it does not satisfy the second. To verify this fact, we first show that if $X$ is a space having a countable basis $\mathscr{B}$, then any discrete subspace $A$ of $X$ must be countable. Choose, for each $a \in A$, a basis element $B_a$ that intersects $A$ in the point $a$ alone. If $a$ and $b$ are distinct points of $A$ the sets $B_a$ and $B_b$ are different, since the first contains $a$ and the second does not. It follows that the map $a \to B_a$ is an injection of $A$ into $\mathscr{B}$, so $A$ must be countable.
I'm having trouble following this. It seems to suggest that $X$ can have any topology with countable basis $\mathscr{B}$ but subspace $A$ is using the discrete topology, and $B_a$ is open in $A$ with the discrete topology. But then it wouldn't make sense that $B_a$ is in $\mathscr{B}$, because $\mathscr{B}$ is a basis for a different topology.
Maybe I'm misunderstanding the term "discrete subspace"? Is that a subspace with the discrete topology? That's what Wikipedia suggests: https://en.wikipedia.org/wiki/Discrete_space
If that's not correct, and the subspace $A$ isn't using the discrete topology, then how can you take an arbitrary point $a \in A$ in the topology of $X$ and choose a basis element $B_a$ that intersects $A$ in the point $a$ alone? For example, if $X=\mathbb{R}, A=(0,2),a=1$, there are basis elements $B_a$ containing $a$, but AFAIK, there is so basis element such that $B_a \cap A = \{a \}$.
Best Answer
The hypothesis says that the topology on $A$ induced from $X$ is the discrete topology of $A$. If $a \in A$ the $\{a\}$ is open in $A$ and so (by definition of subspace topology) there exists an open set $U$ in $X$ such that $\{a\}=U\cap A$.