1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\mathcal{B}$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.