2
$\begingroup$

I want to prove that the subbais of a topology really generates a topology. I think I have been able to prove most of it, but I am not able to show that the empty set is in the topology. Can you please tell me how to show that the empty set is in the topology, and also point out any errors in the other 3 parts if you see it?

The definition of a subbasis in my book is:

A subbasis for $\mathcal{S}$ for a topology X is a collection of subsets of X whose union equals X. The topology generated by the subbasis is defined to be the collection of all unions of finite intersections of elements of $\mathcal{S}$.

I need to prove that it generates a topology. That is, I need to prove that what is generated has this properties:

  1. X is in what is generated.

2. $\emptyset$ is in what is generated.

  1. What is generated is closed under unions.

  2. What is generated is closed under finite intersections.

My attempt:

  1. Since the union of the subbasis equals X this is clear.

2. I do not see how I get the empty set, how do I know that the finite intersections of some sets will be the empty set?

3 and 4:

I have given the set $\mathcal{S}$, I assume it has indexset $I^{\mathcal{S}}$.

I call the set of all finite intersections of $\mathcal{S}$ for $\mathcal{B}$. Each element in $\mathcal{B}$ is $B_k$, where k is in the index-set $I^{\mathcal{B}}$. Each $B_k=\cap_{i \in I^{B_k}}S_i$, where $S_i \in \mathcal{S}$, $I^{B_k} \subset I^{\mathcal{S}}$, $I^{B_k}$ is a finite set.

I call the topology generated(I don't yet know if it is a topology), $\mathcal{T}$, each element is denoted $T_l$, $l \in I^{\mathcal{T}}$. We have by construction that $T_l=\cup_{k \in I^{T_l}}B_k$, where $I^{T_l}\subset I^{\mathcal{B}}$. $I^{T_l}$ need not be finite.

Now to prove 3:

Assume that I have a union of elements in $\mathcal{T}$, let $V=\cup_{l\in I^V}T_l$, $I^V\subset I^{\mathcal{T}}$. I need to show that V is a union of finite intersections of elements in $\mathcal{S}$.

$V=\cup_{l\in I^V}T_l$

$=\cup_{l \in I^V}[\cup_{k \in I^{T_l}}B_k]$

$=\cup_{l \in I^V}[\cup_{k \in I^{T_l}}\{{\cap_{i \in I^{B_k}} S_i}\}]$

$\cup_{(l,k):l \in I^{V}, k \in I^{T_l}}[\cap_{i \in I^{B_k}}S_i]$.

Hence the element is a union of finite intersections of elements in $\mathcal{S}$.

4.

For 4 I will prove it for two elements, and then it will follow by induction.

I have $T_1$ and $T_2$:

$T_1\cap T_2=[\cup_{k_1 \in I^{T_1}}B_{k_1}]\cap[\cup_{k_2 \in I^{T_2}}B_{k_2}]$

$=\cup_{k_1 \in I^{T_1}}[B_{k_1}\cap \{\cup_{k_2 \in I^{T_2}}B_{k_2}\}]$

$=\cup_{k_1 \in I^{T_1}}[\cup_{k_2 \in I^{T_2}}\{B_{k_1}\cap B_{k_2}\}]$

$=\cup_{(k_1,k_2) \in I^{T_1}\times I^{T_2}}[B_{k_1}\cap B_{k_2}]$. And obviously $B_{k_1}\cap B_{k_2}$ is a finite intersections of elements in $\mathcal{S}$ since each B is.

Is this proof correct, and how do I prove point 2?

$\endgroup$
  • 6
    $\begingroup$ The empty set comes in as an "empty union", not as intersection. If $I=\varnothing$ then any set of the form $\cup_{i\in I}S_i$ is empty. $\endgroup$ – drhab Aug 11 '15 at 15:05
  • $\begingroup$ @drhab Thank you, I have a question about the basis of a topology, not subbasis as I asked here. In my book, they prove that for the topology is equal to the collection of all the union of the basis elements, he has another definition of a basis and how it is generated(Munkres).However in the proof with the implication that every element in the topology is written as a union of basis elements he says: "Conversely given $U \in \mathcal{T}$, choose for each $x \in U$ an element $B_x$ of $\mathcal{B}$ such that $x \in B_x \subset U$. Then $U=\cup_{x \in U}B_x$, so U equals a union(cont) $\endgroup$ – user119615 Aug 11 '15 at 15:16
  • $\begingroup$ of elements in $\mathcal{B}$. Must he here also do as you say, take a empty union as you call it?, in order to also get the emptyset? Because U may be empty, and then it is a problem to get the $B_x$? $\endgroup$ – user119615 Aug 11 '15 at 15:17
  • $\begingroup$ If $\varnothing$ is not an element of the basis (as mostly) then: yes. In that case $\varnothing$ can be written as empty union of basiselements. $\endgroup$ – drhab Aug 11 '15 at 15:36
  • $\begingroup$ @drhab Thank you very much! $\endgroup$ – user119615 Aug 11 '15 at 15:38
1
$\begingroup$

A collection $\mathcal{B}$ is a basis of a topology on $X$ if it covers $X$ and: $$\forall P,Q\in\mathcal{B}\forall x\in P\cap Q\exists B_{x}\in\mathcal{B}\left[x\in B_{x}\subseteq P\cap Q\right]$$

Note that this comes to the same as: $$P\cap Q=\cup_{x\in P\cap Q}B_{x}$$ So an intersection can be written as a union.

These condition can shortly be noted as: $$\mathcal{B}^{\stackrel{\bigcap}{f}}\subseteq\mathcal{B}^{\bigcup}$$ where $\mathcal{B}^{\stackrel{\bigcap}{f}}$ denotes the set of finite intersections of elements of $\mathcal{B}$ and $\mathcal{B}^{\bigcup}$ denotes the set of unions of elements of $\mathcal{B}$.

Here $X$ is defined as empty intersection.

Here $\mathcal{B}^{\bigcup}$ is the topology of wich $\mathcal{B}$ serves as base.

If $\mathcal{B}:=\mathcal{S}^{\stackrel{\bigcap}{f}}$ then $\mathcal{B}^{\stackrel{\bigcap}{f}}=\left(\mathcal{S}^{\stackrel{\bigcap}{f}}\right)^{\stackrel{\bigcap}{f}}=\mathcal{S}^{\stackrel{\bigcap}{f}}=\mathcal{B}\subseteq\mathcal{B}^{\bigcup}$

So this shows that $\mathcal{S}^{\stackrel{\bigcap}{f}}$ is indeed a basis, and it is the basis of topology $\mathcal{B}^{\bigcup}=\left(\mathcal{S}^{\stackrel{\bigcap}{f}}\right)^{\bigcup}$.

$\endgroup$

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.