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This is a really basic (no pun intended......no? Ok...) question about what it means to be a basis for a topology.

Here is what I know: If $(X, \mathcal{T})$ is a topological space, and $\mathcal{B} \subseteq \mathcal{T}$ is a basis for $\mathcal{T}$, then we know by definition of basis that the following are true:

  1. For every $x \in X$, $x \in B$ for some $B \in \mathcal{B}$.
  2. If $x \in B_{1} \cap B_{2}$ for some $B_{1}, B_{2} \in \mathcal{B}$, then $\exists B_{3} \in \mathcal{B}$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2}$.

Also, I know that if we have a basis $\mathcal{B}$ for a topology $\mathcal{T}$, then the topology generated by the basis, $\mathcal{T}_{\mathcal{B}}$, consists of all possible unions of elements of $\mathcal{B}$.

Here is my question: I want to prove that if $\mathcal{B}$ is a basis for $\mathcal{T}$, then $\mathcal{T} = \mathcal{T}_{\mathcal{B}}$. Showing $\mathcal{T}_{\mathcal{B}} \subseteq \mathcal{T}$ is really easy. To show $\mathcal{T} \subseteq \mathcal{T_{\mathcal{B}}}$ relies on some "fact" that I don't know to prove: that if $U \in \mathcal{T}$, and $x \in U$, $\exists B \in \mathcal{B}$ such that $x \in B \subseteq U$.

How do I prove that for any $U \in \mathcal{T}$, $\exists B \in \mathcal{B}$ such that $x \in B \subseteq U$? I don't see how this fact follows from the definition of a basis. And without this fact, I can't prove that $\mathcal{T} = \mathcal{T_{\mathcal{B}}}$.

layman
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  • Usually saying $\mathcal{B}$ is a basis for $\mathcal{T}$ means $\langle\mathcal{B}\rangle=\mathcal{T}$ where $\langle\mathcal{A}\rangle:={\bigcup A_\lambda:A_\lambda\in\mathcal{A}}$. Did you mean rather that if $\mathcal{B}\subseteq\mathcal{T}$ satisfies the properties of a basis then $\mathcal{B}$ is indeed a basis for $\mathcal{T}$? – C-star-W-star Sep 08 '14 at 00:23
  • @Freeze_S The way I learned it from the Munkres book, a basis $\mathcal{B}$ for a topology $\mathcal{T}$ is a subset of $\mathcal{T}$ that satisfies properties 1 and 2 in my question. Meanwhile, the topology generated by $\mathcal{B}$ is the set of all unions of basis elements. I just want to show that the topology generated by $\mathcal{B}$ is in fact the same topology that $\mathcal{B}$ is a basis for. – layman Sep 08 '14 at 00:26
  • The "fact" you don't know how to prove appears to have a typographical error: it should end with $x \in B \subseteq U$. That may help, given that this "fact", so corrected, is the exact definition of the statement "$\cal B$ is a basis for $\cal T$". – Lee Mosher Sep 08 '14 at 00:29
  • @LeeMosher Why does $\subseteq$ vs $\subset$ matter? And also, I'm still having trouble tying these two concepts together. As I outlined in my question, I learned one definition for a set $\mathcal{B}$ being a basis for a topology $\mathcal{T}$, and a different definition for what we call the topology generated by the basis elements. And I just want to prove that these two are the same topology. Everyone I ask seems to look at me like I am crazy and stating the same thing twice, but I really don't think I am... – layman Sep 08 '14 at 00:31
  • I'm afraid you might need additionally that if $U\in\mathcal{T}$ then for all $u\in U$ there is $B_u\in\mathcal{B}$ with $u\in B_u\subseteq U$. Otherwise it could happen that ${\varnothing,X}\subseteq\mathcal{T}$ is a basis and generates the indiscrete topology whereas $\mathcal{T}$ is not the indiscrete topology... – C-star-W-star Sep 08 '14 at 00:34
  • @Freeze_S I'm crying internally right now. If we need an additional axiom which doesn't follow from the 2 written above, why isn't it already included in the definition of a basis? – layman Sep 08 '14 at 00:34
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    I will give you a very precise answer in a second - hope that will chear you up again ;) – C-star-W-star Sep 08 '14 at 00:36
  • @LeeMosher I misunderstood which typographical error you were talking about. It's fixed now. – layman Sep 08 '14 at 00:36
  • In other words, the definition of a base as you have given implies that it generates some topology $\mathcal{T}B \subset \mathcal{T}$. For the base to be a base _for $\mathcal{T}$ is exactly the statement $\mathcal{T}_B = \mathcal{T}$, i.e. what you are having difficulty proving. – Zavosh Sep 08 '14 at 01:31
  • Yeah but the problem is that the characterization says nothing about wether the basis is related to some topology. The point is that the basis criterion answers a different problem namely wether it spans a topology at all... Luckily there is also a criterion for the problem wether a collection spans a given topology. Ultimately this criterion is quite different to the characterization of bases... – C-star-W-star Sep 08 '14 at 01:35
  • @Prometheus Yes, you are exactly right. I am having trouble provng $\mathcal{T}_{\mathcal{B}} = \mathcal{T}$ when we know $\mathcal{B}$ is a basis for $\mathcal{T}$. – layman Sep 08 '14 at 02:08

2 Answers2

13

I'm answering my own question because I have a clear answer:

When defining a "basis", we start out with a set $X$. We don't define a topology on it before we define $\mathcal{B}$ axiomatically.

A set of subsets of $X$, $\mathcal{B}$, is said to be a base to the set $X$ if the following two conditions hold:

(i) $\forall x \in X$, $\exists B \in \mathcal{B}$ such that $x \in B$

(ii) Given $B_{1}, B_{2} \in \mathcal{B}$, if $\exists x \in B_{1} \cap B_{2}$, then $\exists B_{3} \in \mathcal{B}$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2}$

Now that we have defined what it means for a set $\mathcal{B}$ of subsets of $X$ to be a base to the set $X$, we can define $\mathcal{T}_{\mathcal{B}}$, the topology generated by $\mathcal{B}$, as the set of all unions of elements in $\mathcal{B}$. That is, a set $U$ is open in $\mathcal{T}_{\mathcal{B}}$ if it is a union of elements of $\mathcal{B}$. It is easy to prove that this is a topology.

Now, if we start out with a topology $\mathcal{T}$ on a set $X$, and we say $\mathcal{B}$ is a basis for the topology $\mathcal{T}$, this is defined as $\mathcal{T}$ actually being the topology $\mathcal{T}_{\mathcal{B}}$, the set of all unions of elements in $\mathcal{B}$.

If $(X, \mathcal{T})$ is a topological space, it is possible to have a set $\mathcal{B}$ of subsets of $X$ satisfy the two properties that make it a base to the set $X$ without it being a basis for a given topology. But if $\mathcal{B} \subseteq \mathcal{T}$, then we are assured $\mathcal{T}_{\mathcal{B}} \subseteq \mathcal{T}$. We don't have equality, though, unless we are also given that $\mathcal{B}$ is a basis for $\mathcal{T}$.

Here is an example of a topological space in which a base to the set $X$ is contained in the topology of $X$ but is not a basis for the topology. Let $(X, \mathcal{T}) = (\mathbb{R}, \mathcal{T}_{\text{indisc}})$ where $\mathcal{T}_{\text{indisc}}$ is the indiscrete topology (i.e., $\mathcal{T}_{\text{indisc}} = \{ X , \emptyset \}$). Then since every topology acts as a basis for itself, $\mathcal{T}_{\text{indisc}}$ is a basis for itself, and it is also a base to the set $X$. However, this base to the set $X$ is contained in $\mathcal{T}_{\text{disc}}$, the discrete topology, but it is not a basis for the discrete topology.

So, the main point here is that when we define a base to the set $X$, it is independent of any topology on $X$. But should the elements of the base be in a topology, then the topology generated by the base is a subset of the original topology. Furthermore, if we say the base is a basis for the original topology, by definition that means the original topology is equal to the topology generated by the base.

layman
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  • Both are called "base" (not "basis"). But if we start with a set $X$ and then consider a collection $\mathcal{B}$, I call $\mathcal{B}$ a "base for a topology" (which is then defined by this $\mathcal{B}$ as said), while if I start with a topology $\mathcal{T}$, and then consider a subfamily $\mathcal{B} \subset \mathcal{T}$ of it, I say that $\mathcal{B}$ is a base for the topology ($\mathcal{T}$). So I think context will make clear which is meant. – Henno Brandsma Sep 08 '14 at 21:09
  • @HennoBrandsma Thanks for your input. Also, I want to make it crystal clear: We can have $\mathcal{B} \subset \mathcal{T}$ satisfy the axioms for $\mathcal{B}$ to be a base to the set $X$, but $\mathcal{B}$ isn't necessarily a basis for $\mathcal{T}$. It is a basis specifically in the case where we have $\mathcal{T}_{\mathcal{B}} = \mathcal{T}$. – layman Sep 08 '14 at 21:26
  • Sure. It's quite uncommon to consider a subfamily of a given topology and call it a base without it generating the original topology. And indeed this is part of what I call a base for the topology. – Henno Brandsma Sep 08 '14 at 21:31
  • @HennoBrandsma But that fact that this scenario is possible should be mentioned because it was a huge source of confusion for me, and it is what caused me to ask this question in the first place. Having a base for the set $X$ doesn't necessarily mean it's a basis for a given topology. I was initially under the impression that it did mean this because of this convention you brought up of only talking about a base under a topology for which it forms a basis. – layman Sep 08 '14 at 21:46
  • @HennoBrandsma: Are you sure there is a difference between base and basis? As far as I know this is only due to different authors but usually not of one author saying I will distinguish between basis and base... As I've seen mostly one will simply say that this collection is a basis for a topology while that collection is a basis for the topology (mentioned, given before, above or whatsoever) but without changing from "base" to "basis"... In the end I would suggest you to rather state things explicitely as $\langle\mathcal{B}\rangle=\mathcal{T}$ rather then describing them in words ;) – C-star-W-star Sep 11 '14 at 12:46
  • @Freeze_S I always use base for both. A basis is for a vector space. I did not suggest using them both in a topological sense. I do think "base" is the more commonly used one. – Henno Brandsma Sep 11 '14 at 14:51
  • @HennoBrandsma: Ah right I'm sorry. But I guess unfortunately user46944 really got the impression base and basis in the context of topological spaces are used differently reading his posts... – C-star-W-star Sep 12 '14 at 00:10
  • @Freeze_S You're right, I don't really understand what it means to be a base. I know if a collection of subsets $B$ is a base for a set $X$, then that means it satisfies properties 1 and 2 in my post above. Now, if we have a set $X$ and a base $\mathcal{B}$ on $X$, then we know that $\mathcal{B}$ is the basis for some topology on $X$ (namely, the topology consisting of all unions of sets in $\mathcal{B}$, i.e., the topology generated by $\mathcal{B}$). $\mathcal{B}$ is then a basis for the topology. That is what I understand about the terms "base" and "basis". – layman Sep 12 '14 at 00:50
  • @user46944: Basis=Base ;) – C-star-W-star Sep 12 '14 at 01:09
  • @Freeze_S I honestly appreciate your time and attempts to help. But unfortunately, the way I learn, telling me "basis = base" when I don't understand why this is true does not help me. Especially after I outlined what I think the difference between the two is. It's similar to our discussion with respect to being a basis for a topology vs a basis for the topology. Unfortunately, telling me repeatedly that something is true doesn't convince me that it is. In math, we should be able to go into as much detail as necessary to prove our claim is true. We don't just say it's true w/o proof. – layman Sep 12 '14 at 02:18
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I just checked Munkres - he made everything fine - but just to clarify:

In principle there are two problems:

a. Given a collection $\mathcal{B}$. Then $\langle\mathcal{B}\rangle$ is a topology iff it satisfies the characteritation: $$\forall x\in X\exists B_x\mathcal{B}:\quad x\in B_x$$ $$\forall B,B'\in\mathcal{B}\forall x\in B\cap B'\exists B_x\in\mathcal{B}:\quad x\in B_x\subseteq B\cap B'$$ b. Given a collection $\mathcal{B}$ and a topology $\mathcal{T}$. Then $\langle\mathcal{B}\rangle=\mathcal{T}$ iff it fulfills the criterion: $$\forall U\in\mathcal{T}\forall u\in U\exists B_u\in\mathcal{B}\subseteq\mathcal{T}: u\in B_u\subseteq U$$

Note that though both problems are conceptually similar they are solved quite differently.

C-star-W-star
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  • I'm still confused as to why this helps me. If we can characterize a basis $\mathcal{B}$ of a topology $\mathcal{T}$ by a collection of elements satisfying 1) $\forall x \in X, \exists B \in \mathcal{B}$ such that $ x \in B$, and 2) if $x \in B_{1} \cap B_{2}$ for basis elements $B_{1}$, $B_{2}$, then $\exists B_{3} \in \mathcal{B}$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2}$, and we can also say that this basis generates a topology of its own, $\mathcal{T}{\mathcal{B}}$, then how do I prove $\mathcal{T} \subseteq \mathcal{T}{\mathcal{B}}$ using the characterizations in the problem? – layman Sep 08 '14 at 02:11
  • From what I saw, Munkres gave these characterizations I listed and then proved $\mathcal{T} = \mathcal{T}_{\mathcal{B}}$ by pulling the following fact out of thin air: if $U \in \mathcal{T}$, then for each $x \in U$, $\exists B \in \mathcal{B}$ such that $x \in B \subseteq U$.... – layman Sep 08 '14 at 02:13
  • Yepp I see... The point is you used sort of answer (a) for question (b). So knowing that 1) [...] and 2) [...] hold doesn't tell ANYTHING!!!! about which topology will come out. – C-star-W-star Sep 08 '14 at 02:17
  • Well that thin air argument is simply wrong ...can you tell me the page? – C-star-W-star Sep 08 '14 at 02:18
  • Sure, Munkres defines a basis as I wrote in 1) on page 78. Then, he defines the topology generated by $\mathcal{B}$ on the same page, which I denote $\mathcal{T}_{\mathcal{B}}$, but he just denotes it $\mathcal{T}$. Then, on page 80, he proves Lemma 13.1, which is exactly what I am trying to prove. Only he pulls the special fact that we are talking about out of thin air. :( – layman Sep 08 '14 at 02:21
  • @user46944: "Topology - A First Course" or "Topology (2nd edition)" – C-star-W-star Sep 08 '14 at 02:28
  • Sorry, it's the 2nd edition! – layman Sep 08 '14 at 02:28
  • @user46944: Ah ok, Munkres defines the topology generated by a basis as the set of all $U$ for which the criterion (b) holds and then finds out that this is the same as taking the span of $\mathcal{B}$. Moreover he proves that this is indeed a topology. – C-star-W-star Sep 08 '14 at 02:36
  • But better do it this way as the name suggests a basis is sort of something that gives you every element as a linear combination that could be interpreted as the unions in other words the span... – C-star-W-star Sep 08 '14 at 02:39
  • So if I am understanding what you are saying correctly, then given a topology $\mathcal{T}$ and a basis $\mathcal{B}$ for the topology $\mathcal{T}$, then Munkres is defining the topology generated by the basis, $\mathcal{T}{\mathcal{B}}$, as ${ U \in \mathcal{T} \mid \forall x \in U, \exists B{x} \in \mathcal{B} \text{ such that } x \in B_{x} \subseteq U }$? – layman Sep 08 '14 at 02:42
  • If this is what he is doing, i.e., defining the topology generated by a basis as a subset of the original topology, then that still doesn't tell me that $\mathcal{T} = \mathcal{T}{\mathcal{B}}$. It only gives me $\mathcal{T} \supseteq \mathcal{T}{\mathcal{B}}$. – layman Sep 08 '14 at 02:43
  • @user46944: No not yet. Given a collection $\mathcal{B}$ define $U\in\mathcal{T}{\mathcal{B}}:\iff\forall u\in U\exists B_u\in\mathcal{B}:u\in B_u\subseteq U$. Since $\mathcal{B}$ satisfies (1) and (2) $\mathcal{T}{\mathcal{B}}$ is indeed a topology. Moreover $U\in\mathcal{T}{\mathcal{B}}\iff U=\bigcup B\lambda,B_\lambda\in\mathcal{B}$. That's it so for but it does not relate yet to any given topology... – C-star-W-star Sep 08 '14 at 02:49
  • @user46944: And yes it only tells you that $\mathcal{T}_{\mathcal{B}}\subseteq\mathcal{T}$. – C-star-W-star Sep 08 '14 at 02:51
  • So best you can do is take the two different approaches (a) and (b)... They are customized to precisely solve the problems at hand... – C-star-W-star Sep 08 '14 at 02:53
  • So Munkres did pull that fact out of thin air. If $\mathcal{B}$ is a basis for a topology $\mathcal{T}$, i.e., it satisfies 1) and 2), then we can't say every element $U$ of $\mathcal{T}$ has the property that $x \in U \implies \exists B \in \mathcal{B}$ with $x \in \mathcal{B} \subseteq U$... – layman Sep 08 '14 at 02:54
  • @user46944: Oh yes we can since the topology he talks about is precisely the one generated by the basis and it is indeed defined precisely by those sets for which this holds... – C-star-W-star Sep 08 '14 at 02:56
  • I don't think so. Read Lemma 13.1 again. He did not say $\mathcal{T}$ is the topology generated by the basis. He just said $\mathcal{B}$ is a basis for the topology $\mathcal{T}$. – layman Sep 08 '14 at 02:58
  • @user46944: These are synonymus. – C-star-W-star Sep 08 '14 at 03:13
  • Haha, this conversation just took a turn for the worse. I'm going to talk to my topology professor tomorrow. – layman Sep 08 '14 at 03:16
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    Okay I just read this discussion and looked in Munkres and it's now clear to me why you're confused. There's a difference between $\mathcal{B}$ being a basis for $\textbf{a}$ topology on $X$ (where $X$ is any set), and a basis for $\textbf{the}$ topology on $X$ (where $X$ is a topological space). The former means that $\mathcal{B}$ satisfies the two conditions for it to generate some topology $\mathcal{T}_\mathcal{B}$ on the set $X$. On the other hand, saying $\mathcal{B}$ is a basis for $\mathbf{the}$ topology on a topological space $X$, that means $\mathcal{B}$ generates the topology of $X$. – Zavosh Sep 09 '14 at 08:06
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    It's strictly a matter of the English language. Munkres only explicitly talks about what it means for $\mathcal{B}$ to be a basis for $\textbf{a}$ topology, and then expects that you understand when we talk about $\textbf{the}$ topology, it's the one it generates. – Zavosh Sep 09 '14 at 08:09
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    By analogy, when $V$ is a vector space and we say $\beta \subset V$ is a basis for a vector space, it simply means $\beta$ is linearly independent. But when we say $\beta$ is a basis for the vector space $V$, it means $\beta$ is linearly independent $\textit{and}$ spans $V$. – Zavosh Sep 09 '14 at 08:14
  • It's exactly the same with topological spaces, and in fact this is the reason it's also called a basis in Topology. – Zavosh Sep 09 '14 at 08:15
  • @Prometheus: Yes, as I said they're synonymous here ;) – C-star-W-star Sep 09 '14 at 14:09
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    @Freeze_S: Yes, when I said 'you' obviously I meant user46944. I also did say the two things were synonymous in perhaps less clear words the day before, but that didn't appear to help user46944. One can't easily resolve a confusion by simply overruling it. The source of the confusion has be neutralized first. – Zavosh Sep 09 '14 at 15:42
  • @Prometheus: Yes you're right :) and as you correctly observed it is (again) a matter of language... – C-star-W-star Sep 09 '14 at 16:14
  • @Prometheus Where were you two days ago??? Lol but yes you are exactly right about where I was confused. Also, you gave an excellent analogy about the basis of some vector space and the basis of the vector space. Seriously, a lot of people don't have the ability to understand where a person's confusion comes from, so I don't even know you, but I already look up to you. I'll bet you're an awesome teacher. – layman Sep 11 '14 at 00:20
  • @Prometheus By the way, I haven't chosen my answer as a best answer yet. If you decide you want the extra recognition on the site (I'd say you deserve it), add your answer and analogy as a post and I'll gladly accept your answer as best. – layman Sep 11 '14 at 00:27
  • @user46944: No need for extra credit. Glad to be of help :) – Zavosh Sep 11 '14 at 00:34
  • @layman If Zavosh is correct you should accept his answer, because your accepted answer, where you differentiate between base and basis is confusing in light of these comments, which appear to say they are one and the same, and that from then on it is just a matter of specifying 'a basis' or 'the basis' when talking about generating topologies. – ManUtdBloke Aug 22 '17 at 16:25
  • @eurocoder Zavosh is absolutely correct in the comments. However, he/she did not post an answer, so I cannot accept his/her answer. The answer under which all of these comments lie is for user Alexander Frei, previously named Freeze_S, and I still remember the great confusion he caused me when I first asked this question. The answer I posted will remain the accepted answer as it totally clears up the confusion I had when asking this question. – layman Aug 22 '17 at 21:21
  • @layman It sounds from your answer as if 'base' and 'basis' are two different entities. But then in these comments it is accepted that they are interchangeable terms for the same thing, unless I am missing something. We can take of a 'base for a topology' when not being specific and a 'base for the topology' when being specific - or equivalently speak of a 'basis for a topology' when not being specific and a 'basis for the topology' when being specific. So mixing the terms makes your post confusing! – ManUtdBloke Aug 23 '17 at 10:33
  • @eurocoder Basis and base for a topology really are the same that is what I was trying to say. It always means $\langle\mathcal{B}\rangle=\mathcal{T}$. The thing is just that not for every family $\mathcal{S}\subseteq\mathcal{P}(X)$, $\langle\mathcal{S}\rangle$ is a topology. – C-star-W-star Aug 23 '17 at 13:36
  • @eurocoder Nowhere in my answer do I use the word "base" in seclusion. I'm always using it in the phrase "base to the set", and this is what I define in the answer. I define what is means for a collection of subsets of $X$ to be a "base to the set $X$" for some set $X$. I don't know that you can rename a "base to the set $X$" as a "basis to the set $X$", but it's been a long time since I've looked at the source material for this stuff. I reread my answer, and don't find it confusing given the information I've just presented. – layman Aug 23 '17 at 17:09