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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 $\mathcal{T} = \mathcal{T_{\mathcal{B}}}$.

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  • $\begingroup$ 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 $\mathcal{T}$? $\endgroup$ – C-Star-W-Star Sep 8 '14 at 0:23
  • $\begingroup$ @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. $\endgroup$ – layman Sep 8 '14 at 0:26
  • $\begingroup$ 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 $\cal T$". $\endgroup$ – Lee Mosher Sep 8 '14 at 0:29
  • $\begingroup$ @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... $\endgroup$ – layman Sep 8 '14 at 0:31
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    $\begingroup$ I will give you a very precise answer in a second - hope that will chear you up again ;) $\endgroup$ – C-Star-W-Star Sep 8 '14 at 0:36
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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.

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  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Sep 8 '14 at 21:09
  • $\begingroup$ @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 $\mathcal{T}_{\mathcal{B}} = \mathcal{T}$. $\endgroup$ – layman Sep 8 '14 at 21:26
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Sep 8 '14 at 21:31
  • $\begingroup$ @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. $\endgroup$ – layman Sep 8 '14 at 21:46
  • $\begingroup$ @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 ;) $\endgroup$ – C-Star-W-Star Sep 11 '14 at 12:46
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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.

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  • $\begingroup$ 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? $\endgroup$ – layman Sep 8 '14 at 2:11
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    $\begingroup$ 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 $X$. $\endgroup$ – Zavosh Sep 9 '14 at 8:06
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    $\begingroup$ 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. $\endgroup$ – Zavosh Sep 9 '14 at 8:09
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    $\begingroup$ 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 $V$. $\endgroup$ – Zavosh Sep 9 '14 at 8:14
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    $\begingroup$ @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. $\endgroup$ – Zavosh Sep 9 '14 at 15:42

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