Confused about the definitions of cofinite and cocountable topologies Note: $cc$ stands for "cocountable" and $cf$ stands for "cofinite".
Statement:

(i) Any one-to-one map from $(\mathbb R, \tau_{cc})$ to $(\mathbb R, \tau_{cf})$ is continuous.


(ii) Any continuous map from $(\mathbb R, \tau_{cf})$ to $(\mathbb R, \tau_{usual})$ is constant.

Proof:

(i) Let $F$ be $\tau_{cf}$ closed. $\color{red}{\text{Then either $F = \mathbb R$ or $F$ is finite.}}$ Thus, in the first case, we have $F = \mathbb R \implies f^{-1}(F) = \mathbb R \implies  f^{-1}(F)$ is $\tau_{cc}$ closed and in the latter case, $f$ is $1-1$ and $f$ is finite $\implies f^{-1}(F)$ is finite $\implies f^{-1}(F)$ is countable $\implies \color{green}{\text{$f^{-1}(F)$ is $\tau_{cc}$ closed}}$. Hence $f$ is continuous.


(ii) Suppose $g$ is not constant. Then exist $x, y \in \mathbb R$ s.t. $g(x) > g(y).$ Pick disjoint, open intervals $I, J$ in $\mathbb R$ s.t. $g(x) \in I, g(y) \in J$. Now $g^{-1}(I) \cap g^{-1}(J) = g^{-1}(I \cap J) = g^{-1}(\emptyset) = \emptyset$ and $x \in g^{1}(I), y \in g^{-1}(J)$, so $g^{-1}(I), g^{-1}(J)$ are nonempty, disjoint and $\color{blue}{\text{$\tau_{cf}$-open}}$. But $\mathbb R = \mathbb R \setminus \emptyset =g^{-1}(I) \cup \mathbb R \setminus g^{-1}(J) = $ union of two finite sets. Contradiction.

My question:

*

*$\color{red}{\text{part in red}}: \ F \in \tau_{cf} \text{ and $F$ is finite $\implies \mathbb R \setminus F$ is finite and $\mathbb (\mathbb R \setminus F) \cup F = \mathbb R =$ finite}$. Looks like a contradiction to me.

*$\color{green}{\text{part in green}}:$ The expression in green implies $\mathbb R \setminus f^{-1}(F)$ is countable. How do we know this subset of reals is countable? It's treated as something self-evident, but it's not obvious to me.

*$\color{blue}{\text{part in blue}}:$ The expression in blue implies $\mathbb R \setminus g^{-1}(I)$ is finite. How do we know this? Again, not at all obvious to me :)

Can someone, please, comment on these? Thanks.
 A: $\color{red}{\text{red: } } $The definitions do have these exceptions or else they would not be topologies:

*

*In the cofinite topology of a set $X$, the closed sets are precisely the finite sets -- as well as $X$ itself (because in every topology, $\emptyset$ and $X$ are closed)

*In the cocountable topology of a set $X$, the closed sets are precisely the countable sets -- as well as $X$ itself (because in every topology, $\emptyset$ and $X$ are closed)

$\color{green}{\text{green: } } $Note that the argument uses closed sets, not open sets, hence finite/countable sets, not cofinite/cocountable sets.
$\color{blue}{\text{blue: } } $ Open follows from continuity.
A: The red part is not a contradiction, it's the definition of the cofinite topology: the only closed sets are the finite ones (including $\emptyset$) and $\Bbb R$ (because it has to be closed in any topology). The open sets are the complements of these. So if $F$ is closed, it's not the case that $\Bbb R\setminus F$ is finite, but $\Bbb R\setminus F$ is open, but the proof doesn't want to discuss open sets, but show continuity just using closed sets.
In green we just use that that a finite set is countable so closed in $T_{cc}$.
in blue: the inverse image of open is open (by continuity) and the only non-empty open sets in the cofinite topology are the ones with a finite complement.
