Fake proof of $\mathbb{R}$ being countable

Question

We know that $\mathbb{R}$ is second countable. Let $\{V_n\}$ be a countable basis. To every $a\in \mathbb{R}$ consider the interval $(-\infty,a)$. Now find the least $n$ such that $(-\infty,a)^c\cap V_n\neq\phi$ and $V_n\subset(-\infty,b)$ for all $b>a$. Now map $n\rightarrow a$. This is one-to-one. Can anyone tell me where I am going wrong? Thank you.

Answer 1

If an open set intersects $[a, \infty)$, then it is not contained in all $(-\infty, b)$ for $b > a$. Thus, the set of indices you want to take the minimum of is empty.

Answer 2

An error is that you've left out the final step: you haven't even concluded that $\mathbf{R}$ is countable!

Your final statement is only that an injective map exists, not a bijective map, and thus would only imply that the cardinality of $\mathbf{R}$ is at least the cardinality of $\mathbf{N}$.

Answer 3

@Arno is completely correct.

Your problem seems to be related to the properties of a basis. Suppose that $\mathcal{V}=\{V_{n}\vert n\in\mathbb{N}\}$ is our countable basis. We know that every open set $U$ is the union of the basis elements it contains, and vice versa that the union of any collection of basis elements is also open. (Additionally, we know that the intersection of a finite collection of basis elements is also in the basis, but that isn't so relevant here).

So, what are you trying to do? Given a real number $a$, you want to find a basis element (i.e., an open set) $V$ which intersects the complement of $(-\infty, a)$ and is contained in $$\bigcap_{b>a}(-\infty,b).$$

Your main problem is that big intersection. I think that you may be getting confused about the intersection of the $(-\infty,b)$ sets. The intersection of a finite number of open sets is necessarily open, but this is the intersection of an infinite family. Thus it is not true that this intersection can be written as a union of elements of $\mathcal{V}$. In particular, it is not the case that $a$ lies in the interior of this intersection. Thus, no element of $\mathcal{V}$ which contains $a$ lies in this intersection.

In fact, if you think about it, this intersection is just $(-\infty,a]$.

---

Also, the way you wrote $[a,\infty)$ as $$(-\infty,a)^{c}$$ makes me suspect that you temporarily forgot that the complements of open sets aren't necessarily open.