# A few questions about topological spaces

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 $$f^{-1}(\mathbb R) = \mathbb R?$$

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.

(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.

• I haven't carefully read your question, but I'd highly recommend deleting everything below the horizontal line, and posting it as a new question. Shorter, more focussed questions will gather more answers, while longer questions with a large number of parts may be closed for being too broad. Commented Jul 4, 2021 at 20:35

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).

• Let me see if I understand you correctly. All the elements $X$ in the co-countable topology $\tau_{cc}$ are open by definition of topology. No countable set $X^c$ belongs in $\tau_{cc}$ by definition of $\tau_{cc}.$ Then since closed set is defined as the complement of an open set, $X^c$ must be closed. As reals are both open and closed, the only closed element in $\tau_{cc}$ must be $\mathbb R.$ Since $f^{-1}(\{y\}) \ne \mathbb R$ and $f^{-1}(\{y\})$ is closed, $f^{-1}(\{y\}) \not \in \tau_{cc}$ meaning it's countable. Commented Jul 4, 2021 at 22:07
• @user947160 No, not quite. If an element is not in $\tau_{cc}$ it can be countable or uncountable. We only know it's not $\emptyset, \Bbb R$ and its complement is not countable. E.g. a set like $(0,1) \subseteq \Bbb R$ is not in the topology, as its complement is uncountable. A set $C$ is countable iff $\Bbb R\setminus C$ has countable complement (trivially) iff $\Bbb R\setminus C$ is open so in $\tau_{cc}$ iff $C$ is closed. The only other closed set is the complement of $\emptyset$, i.e. $\Bbb R$ itself. Commented Jul 5, 2021 at 12:11
• @user947160 WE often define the topology $\tau_{cc}$ as $C$ is closed iff $C=\Bbb R$ or $C$ is at most countable. Commented Jul 5, 2021 at 12:12