Countable topology on $[0,1]$

Question

Problem 4.25

Let T be the collection of all subsets $U \subset [0, 1]$ such that $[0, 1] − U $ is countable, together with the empty set.

1. Prove that T is a topology on $[0, 1]$.

$\emptyset\in \mathcal T$ and if we delete no points from $[0,1]$ then $X\in\mathcal T$.

Let $\cup_\alpha U_\alpha$ be open sets, then $U_\alpha=X-G_\alpha$ where $F_\alpha$ is at most countable.

So we have that $\cup_\alpha U_\alpha=\cup (X-F_\alpha)=X-\cap_\alpha F_\alpha$, however $\cap_{\alpha} F_\alpha \subset F_\alpha$ is countable so topology.

2. Find the closure of $(0, 1]$.

Seems to just be $[0,1]$ quite honestly i was too lazy to prove this as it is obvious clearly it is not closed and it is a subset of $[0,1]$ consider if $(0, 1] \subset [a,1]$ where $a>0 $ assume this is the smallest closed set, consider $[\frac{a}{2},1] $ since $a>0$ we have that $\frac {a}{2} >0 $ this is a contradiction so leads us to $ a\leq 0 $ but $a<0$ is nonsense.

3. Prove that there is no sequence in $(0, 1]$ which converges to $0$.

i did eventually get this question and i still believe posted answer is wrong. $U=[0,1] / \{x_1,x_2,... \}$ then $ U^c =\{x_1,x_2,... \} $ is countable so $U\in \mathcal T $ but U does not contain any infinite tail of the sequence! so not only does the sequence not converge to 0 it appears that no sequence converges to anything.

Edit: This question has correct solutions to all three part of the problem (not just attempts) not sure how it was closed for lack of information given by OP

Answer 1

2. The answer is $\mathbb{R}$. Any uncountable subset will indeed be dense, as an open set can only exclude countably many points, and thus must contain at least some points in the uncountable subset.

3. Suppose $(x_n) \in \mathbb{R}$ such that $x_n \to 0$ in under the co-countable topology, but $x_n \neq 0$ for all $n$. Then the range of $(x_n)$ is countable, so its complement is open. This is an open set, containing $0$, which fails to intersect the sequence. But, if $x_n \to 0$, then infinitely many terms of $x_n$ should lie in every open neighbourhood of $0$, which is a contradiction.