The ''Countable Complement Extension Topology'' $\tau$ on $\mathbb{R}$ is the topology whose open sets of the form $$\widetilde{U}=U\setminus A,$$ where $U$ is a standard open set and $A$ is some countable set. The terminology borrowed from the very popular book Counterexamples in topology. I could use some help to show that arbitrary unions of open sets is open (which is needed to check that this defines a topology).

My attempt: For me it is easier to look at this problem from the closed sets point of view. Closed sets of $\tau$ are of the form $$V=\widetilde{V}\cup A$$ for some usual closed set $V$ and countable $A$, and I want to check that arbitrary intersections look again a closed union a countable. Let write $\bigcap_{\gamma\in\Gamma} (V_\gamma\cup A_\gamma)$. Now certainly $A_\gamma\cup L_\gamma=\overline{A_\gamma}$, where $L_\gamma$ is the set of limit points of $L_\gamma$, which will by countable. Now $(V_\gamma\cup \overline{A_\gamma})$ is closed, so $\bigcap_\gamma (V_\gamma\cup \overline{A_\gamma})$ is closed. And moreover there exists some set $\Delta\subset L_1$ such that $$\bigcap_\gamma (V_\gamma\cup A_\gamma)\uplus\Delta=\bigcap_\gamma (V_\gamma\cup \overline{A_\gamma}),$$ and therefore the result follows.

Is this correct? Is there an alternative/easier way to show that the union of open sets is open?

Thanks in advance for your time,

Some background: In the (aforementioned) book $\tau$ is defined as the topology generated by $\tau_1\cup\tau_2$, $\tau_1$ being the usual topology and $\tau_2$ the countable complement topology. From there I can show that sets of the form $U\setminus A\;\;$ ($U$, $A$, as before) are indeed open in $\tau$, and form a base of $\tau$, but I am having difficulty seeing that the most general open looks like that.

EDIT: For completeness, I add here a sketch of the proof that arbitrary unions of elements in $\tau$ look again like $\widetilde{U}$. To do that, consider a counatble base $\{(a_n,b_n)\}_n$ of the standard topology, and use it it to write each of the $\widetilde{U}_\gamma$. Then express $\cup_\gamma \widetilde{U}_\gamma=\cup_\gamma U_\gamma\setminus A_\gamma$ as $\cup_{n\in J} (a_n,b_n)\setminus A_n$, where $J\subset\mathbb{N}$ is the index that tells you whether $(a_n,b_n)$ is contained in some $\widetilde{U}_\gamma$, and $A_n$ is constructed as intersection of the $A_\gamma$ (precisely for those $\gamma$ such that $(a_n,b_n)$ is contained in $U_\gamma$). From here, the result follows easily since $\cup_n A_n$ will be countable.

  • $\begingroup$ You could use de Morgan laws, use that arbitrary intersection of countable sets is countable $\endgroup$ – Mirko May 22 '17 at 3:44
  • $\begingroup$ I don't see why you need $\tau$ be generated by $\tau_1 \bigcup \tau_2$, since for any open set $U$, let $A=\varnothing$, then $\widetilde{U} = U$ and thus $\tau_1 \subset \tau_2$, so $\tau = \tau_2$. $\endgroup$ – MaudPieTheRocktorate May 22 '17 at 4:30
  • $\begingroup$ That's not correct: $\tau_1$ is not contained in $\tau_2$. Opens in $\tau_2$ are complements of countable sets. But then $O=(0,1)\in\tau_1$ is not open in $\tau_2$. You certainly have $\tau_2\subset\tau$, but the converse is not true, and you might use $O\in\tau$ again as a counterexample. $\endgroup$ – Simulacra May 22 '17 at 11:37

The final remarks are more to the point. Indeed if $\tau_1$ is the usual topology and $\tau_2$ the co-countable one on $\mathbb{R}$, then we define the topology $\tau = \tau_1 \lor \tau_2$ (the sup in the lattice of topologies) as the smallest topology that contains $\mathcal{S} = \tau_1 \cup \tau_2$. And by definition $\mathcal{S}$ is a subbase for $\tau$ and a standard theorem then learns us that the finite intersections from $\mathcal{S}$ are a base for $\tau$.

What do intersections $O_1 \cap \ldots O_n \cap U_i \cap \ldots U_m$ look like, where the $O_i$ are from $\tau_1$ and the $U_i$ from $\tau_2$? As topologies are themselves closed under finite intersections we get all sets of the form $O \cap U$ where $O \in \tau_1, U \in \tau_2$. A non-typical member of $\tau_2$ is $\mathbb{R}$ or $\emptyset$ which results in either $O$ or $\emptyset$ as the intersection. A more typical set from $\tau_2$ is of the form $U =\mathbb{R}\setminus A$, $A$ countable and then $O \cap U = O \setminus A$. (This includes all members of $\tau_1$ when we take $A = \emptyset$ and all members of $\tau_2$ if we take $O = \mathbb{R}$). So the basis (it is a basis for a topology as it covers $\mathbb{R}$ and is by definition closed under finite intersections) is $$\mathcal{B} = \{O \setminus A: O \text { Euclidean open }, A \text{ at most countable }\}$$

So all open sets are just unions from families from $\mathcal{B}$. All relevant questions about a topology (like continuity ,connectedness ,compactness) can be decided just by knowing the elements from a base. Exercise: find an explicit open set $O$ not from $\mathcal{B}$.

  • $\begingroup$ Thanks for the remarks Henno. As mentioned in the original question, I could indeed find the base $\mathcal{B}$ you are showing. My main difficulty was showing that general open sets had the form $O\setminus A$ as well. $\endgroup$ – Simulacra May 23 '17 at 11:56
  • $\begingroup$ Should my edit be right, there is no open set off $\mathcal{B}$. $\endgroup$ – Simulacra Jun 13 '17 at 20:30

Counterexample: Consider the topology on $\mathbb{R}^2$: A set $U$ is open iff $\forall x\in\mathbb{R}, \{y\in\mathbb{R}: (x,y)\in U\}$ is open. In this topology, the open sets are made by "vertical slices of open sets of $\mathbb{R}$".

Now, let $U_x=\{x\}\times\mathbb{R}=\{(x,y)\in\mathbb{R}^2: y\in\mathbb{R}\}$, $A_x=\{(x,1),(x,1/2),(x,1/3),(x,1/4),\cdots\}$, then $$\bigcup_{x\in\mathbb{R}}(U_x\setminus A_x)=\mathbb{R}\times(\mathbb{R}\setminus\{1, 1/2, 1/3, 1/4, ...\})$$ cannot be expressed as $U\setminus A$, where $U$ is an open set, and $A$ is a countable set.

Suppose it can, then since $\mathbb{R}\setminus\{1, 1/2, 1/3, 1/4, ...\}$ is not open, the section of $U$ at any $x$ must be bigger than that, that is, $$\forall x\in\mathbb{R}, \mathbb{R}\setminus\{1, 1/2, 1/3, 1/4, ...\}\subsetneqq\{y\in\mathbb{R}: (x,y)\in U\}$$ so $\forall x\in\mathbb{R},\exists y\in\mathbb{R}, (x,y)\in A$, thus $A$ must be uncountable.


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