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Im kinda new studying topology, and one of the concepts I find hard to fully understand are basis. From the definition $B$ is a basis if for each open set $O$ in the given topology and each point $x$ of it, there exists a $b$ subset of $B$, such that $x$ its contain in $b$ and as wall as in $O$.

The topology given is: Let $A\subset X$, and $\tau:=\{X\} \cup \{O \subset X : O \cap A = \emptyset\}$

I proceed to prove that $\tau$ is a topology.

It's clear that $X \in \tau $, and $\emptyset \in \tau$ since the intersection with is always empty.

Let, $O_1,O_2 \in \tau$ it implies that $ O_1\cap A = \emptyset$ and $O_2\cap A = \emptyset$ Now, to prove that their intersection is in $\tau$ then $ (O_1 \cap O_2) \cap A = \emptyset$ but by the distributivity of the intersection it follow $ (O_1 \cap O_2) \cap A = (O_1 \cap A) \cap (O_2 \cap A)$ which because they are open, it's empty, so $(O_1 \cap O_2) \in \tau$

To prove the final, let $\{O_n\}_{n\in I} \in \tau$, so $\forall n \in I$ $O_n \in \tau$, further more, $O_n \cap A =\emptyset$. To prove that $\bigcup_{n \in I} O_n \in \tau$, then $ \bigcup_{n \in I} O_n \cap A =\emptyset $ which is true because $ \bigcup_{n \in I} O_n \cap A = \bigcup_{n \in I} (O_n \cap A)$ and since for each $n$ that instersection is empty, so the union must be empty as well.

To find a basis (not trivial) I have to construct a family of sets such that, there is an open set that contains them.

I say that $B=\{u\subset X : u \subset A\}$ is a basis for $\tau$, it's the only collection that I can think off.

But to show that indeed is a basis I don't see clear that I can find such a set $b \in B$. Further more, that $\forall O \in \tau$, and $\forall x \in O$, $\exists b \in B$ such that $x\in b \subset U$. Even more, how to prove this.

Any help in constructing a basis and tips are welcome.
Thank you in advance.

Also, I'm truly sorry If there is a mistake in the spelling or the grammar.

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  • $\begingroup$ I've seen that the set B is not a basis because there is no open set that can be seen as the union of some sets of B. An open set in the topology is those sets which don't have any elements in common with A, so the union are sets of subsets of A cannot form an open set. But If I take the set $B=\{ U \subset X : U_i \cup U_j = \emptyset , i \neq j \}$ I think is indeed a basis. $\endgroup$ – Lilian Hernández Mar 30 '18 at 4:15
  • $\begingroup$ Your definition of a base is incorrect. $\endgroup$ – William Elliot Mar 30 '18 at 4:40
  • $\begingroup$ The B you gave is not a base because not any empty subset U of A does not have empty U $\cap$ A. $\endgroup$ – William Elliot Mar 30 '18 at 4:46
  • $\begingroup$ @WilliamElliot Which one? the one of subsets of A?, in the previous comment I said why (I think) it isn't. Thank you. $\endgroup$ – Lilian Hernández Mar 30 '18 at 5:14
  • $\begingroup$ @WilliamElliot, why it isn't the correct def ? Am I missing something? $\endgroup$ – Lilian Hernández Mar 30 '18 at 5:15
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That topology is called the exclude A topology.
The smallest base is { {x}, X : x not in A }.

Exercise. What is the closure of {a} if a in A?

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  • $\begingroup$ I think... not sure $\bar{ \{a\} } =A $. Am I right? $\endgroup$ – Lilian Hernández Mar 30 '18 at 5:27
  • $\begingroup$ No, it is X. Perhaps tomorrow night I will post a second answer addressing your other conserns. $\endgroup$ – William Elliot Mar 30 '18 at 6:33
  • $\begingroup$ Thanks!, but one question tho. We know that X is open in the topology, and if the B you propose is a base, then X should be equal the union of basic sets, but $A \in X$. ¿Is still true that X can be seen as the union of singletons even tho those do not contain the points in A? $\endgroup$ – Lilian Hernández Apr 2 '18 at 3:22
  • $\begingroup$ @LilianHernández. Note that I included X in B. $\endgroup$ – William Elliot Apr 2 '18 at 8:53
  • $\begingroup$ Sorry @William Elliot, I thought that was a mistake and meant a $\in$. Now everything makes sense. Thank you so much. $\endgroup$ – Lilian Hernández May 11 '18 at 6:50

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