A few questions about topological spaces

Question

Statement $1$: Given a non-constant map $f: (\mathbb R, \tau_{cocountable}) \to (\mathbb R, \tau_{cofinite})$, show $f$ is continuous iff, for each $y \in \mathbb R$, the set $f ^{–1}(\{y\})$ is countable

Proof of Statement $1$: Suppose $f$ is continuous. For each $y \in \mathbb R, \ \{y\}$ is $\tau_{cofinite}$-closed, so $f^{–1}(\{y\})$ must be $\tau_{cocountable}$-closed ($\ne \mathbb R$ since $f$ is non-constant), therefore $f^{–1}(\{y\})$ is countable. Suppose now $f^{–1}(\{y\})$ is countable for each $y \in \mathbb R$. For any closed $K \in (\mathbb R, \tau_{cofinite}), \ K$ is either $\mathbb R$ or finite. Now $f^{–1}(\mathbb R) = \mathbb R$ is $\tau_{cocountable}$-closed, and if $K$ is finite, then $f^{–1}(K) = \bigcup_{y \in K}f ^{–1}(\{y\})$ is a finite union of countable sets, therefore countable, therefore $\tau_{cocountable}$-closed again and hence $f$ is continuous.

My Questions about the proof above:

How does the bolded "therefore" above follow? By definition, any(?) subset $X$ in the domain of $f$ is of the form $\mathbb R = X \cup X^c$ where $X^c$ is the countable complement of $X$. But that makes $\mathbb R$ countable.

In the other direction, how does the italicized finite follow? If $K$ is finite, then $\mathbb R = K \cup K^c$ implies $\mathbb R$ is finite.

In general, does $\tau_{cocountable}$ contain countable sets and does $\tau_{cofinite}$ contain finite sets? And maybe trivially, how do we know $f^{-1}(\mathbb R) = \mathbb R?$

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Question $2$:

Are the following pairs of spaces homeomorphic?

(i) $\mathbb Q \cup (0, 1)$ and $\mathbb R \setminus \mathbb Q$ (as subspaces of $(\mathbb R, \tau_{usual}))$;

(ii) $(\mathbb C, \tau_{disc})$ and $(\mathbb C, \tau_{cocountable})$;

(iii) a circle with one point removed (in $\mathbb R^2$ with $\tau_{usual}$) and $(\mathbb R, \tau_{usual})$;

(iv) $\mathbb N$ with topology $\{1, 2\}, \{3, 4\}, \{5, 6\}, \{7, 8\}, \ldots$ and all unions of these, and $\mathbb N$ with topology $\{1, 3\}, \{2, 4\}, \{5, 7\}, \{6, 8\}, \{9, 11\}, \{10, 12\}, \ldots$ and all unions of these.

The following have answers

(v) $(\mathbb R, \tau_{cocountable}), \ (\mathbb R, \tau_{cofinite})$

(vi) $\mathbb R^2$ and the surface of a sphere with one point removed (natural metric topologies here)

(vii) $(\mathbb Q,l_x), (\mathbb Q, l_y)$, where $x, y$ are two distinct rational numbers and $l_p$ is included-point topology

Book's answers to (v) and (vii):

(v) No. The statement ‘there is a countably infinite closed subset’ is (obviously) a homeomorphic invariant, is true in $(\mathbb R, \tau_{cocountable})$ and is false in $(\mathbb R, \tau_{cofinite})$

(vii) Yes. The map $h : \mathbb Q → \mathbb Q$ given by $h(x) = y, h(y) = x, h(z) = z$ when $z \ne x$ or $y$ is routinely checkable to be a homeomorphism.

My answers/questions regarding question $2$ above:

(i) $\mathbb Q, (0, 1)$ are not compact and so their union is not compact meaning there's no continuous inverse function from $\mathbb Q \cup (0, 1)$

(ii) Any function from a discrete space to an indiscrete space is continuous. Since idenity $i$ is continuous and bijective, $i$ should work here to show homeomorphism

(iii) I think this is a special case of Stereographic projection

(iv) Any inverse image of an open set in $\mathbb N$ is a union of $2$-sets. Since the given $2$-sets are open, their union is open as well. This works in both directions, so a continuous function with its continuous inverse likely exists. I am not sure how to find a concrete homeomorphism, but would $f(x, y) = (x, y – 1)$ if $y$ is odd and $f(x, y) = (x + 1, y)$ if $x$ is even work?

Are my answers to (i) through (iv) correct? If not (or if incomplete), how do I improve them?

(v) This question might be trivial, but I am still new to the very basics of topology. Consider $f: (\mathbb R, \tau_{cofinite}) \to (\mathbb R, \tau_{cocountable})$. Every $Y \in (\mathbb R, \tau_{cocountable})$ must have a pre-image $X \in (\mathbb R, \tau_{cofinite})$. By definition, every $X$ is open, but for $f$ to be continuous $X$ must be cocountable. Correct?

(vi) It's yet another special case of Stereographic projection. Correct?

(vii) Is $h(y)$ a typo? Did they mean $h^{-1}(y)?$

Thanks.

Answer 1

I'll only get into 1, for focus:

In the co-countable topology the only closed sets are $\Bbb R$ and all countable sets (including all finite subsets and also $\emptyset$). As $f$ is not constant, $f^{-1}[\{y\}]$ is ruled out (or $f$ would have been constant with value $y$).

$f$ is continuous iff the inverse image of every closed set is closed. So the condition is necessary as indeed all sets $\{y\}$ are closed in the cofinite topology.

It's sufficient as all closed sets of the cofinite topology are $\Bbb R$ and all finite subsets. $f^{-1}[\Bbb R]=\Bbb R$ (this just says that any $x \in \Bbb R$ is mapped into $\Bbb R$ under $f$, which is a given) and this is closed in any topology on $\Bbb R$. A finite set is a finite union of singletons and the condition on $f$ tells us that all singletons have closed inverse images and as $$f^{-1}[\{y_1,\ldots, y_n\}]=\bigcup_{i=1}^n f^{-1}[\{y_i\}]$$ and trivially also $f^{-1}[\emptyset]=\emptyset$ we see that all inverse images of closed sets are closed and thus $f$ is continuous.

The cofinite topology on $\Bbb R$ does not contain any finite set except $\emptyset$, as $\Bbb R$ is infinite.

The cocountable topology on $\Bbb R$ does not contain any countable set except $\emptyset$, as $\Bbb R$ is uncountable.

(recall that a closed set is defined as the complement of an open set).