2

I am having some trouble understanding the definition of a base, according to wikipedia.

According to wikipedia, a base is defined as follows:

A base is a collection B of subsets of X satisfying these two properties:

1) The base elements cover X. 2) Let B1, B2 be base elements and let I be their intersection. Then for each x in I, there is a base element B3 containing x and contained in I.

I thought I understood this but then wikipedia states the following sentence:

If a collection B of subsets of X fails to satisfy either of these, then it is not a base for any topology on X.

Wikipedia's definition defines a base for a set X. So what do they mean by 'a base for a topology on X'? Say we have a topological space (X,T), what does a base B have to do with T? My understanding of the definition of base leads me to believe that B is fully defined with respects to X.

Pierre
  • 121

3 Answers3

2

The main point you may be missing is that there can be more than one topology for $X$. For example, if $X=\{1,2\}$, then one possible topology on $X$ is $\tau_1 = \{\emptyset, \{1,2\}$, while another topology is $\tau_2=\{\emptyset, \{1\},\{2\}, \{1,2\}\}$.

Note that the two topologies are not the same, and, for example, while $B=\{\{1\}, \{2\}\}$ is a basis for $\tau_2$, it is not the basis for $\tau_1$.

So, what the sentence

If a collection B of subsets of X fails to satisfy either of these, then it is not a base for any topology on X.

is trying to (rather awkwardly) say is that there exists no topology on $X$ for which $B$ is a basis.

5xum
  • 123,496
  • 6
  • 128
  • 204
  • I'm sorry, I don't mean to be slow, I just don't understand. In my head, once you have told me what B and X are, I've got all I need to check whether B is a base or not. (1) I check that the elements of B cover X. Here X = the union of {1} and {2}. So that checks out. (2) I check that any element in the intersection I of any two elements of B is contained in some element of B that is a subset of I. Here any intersection I is the empty set. So that checks out. Clearly I've misunderstood something but I don't see what. – Pierre Mar 28 '17 at 12:48
  • @Pierre Yes, you have now proven that $B$ is a basis. But the question now is, "what topology is it a basis for"? And the answer, in this case, is that it is a basis for $\tau_2$. $B$ is not the basis for $\tau_1$, on the other hand, because $B$ is not a subset of $\tau_1$. – 5xum Mar 28 '17 at 12:49
  • So the point is that there are two distinct definitions in play here: (1) B is a basis (which depends only on X) (2) B is a basis for some topology T which requires that (a) B be a basis (with respects to X) and (b) T = the set of countable unions of elements from B. Is that right? – Pierre Mar 28 '17 at 13:09
  • 1
    @Pierre No, there's just one definition, and that is that $B$ is a basis. Saying "$B$ is a basis" is the same as saying "$B$ is a basis for some topology". And if $B$ satisfies the two conditions you wrote down, then $B$ is a basis (for some topology). – 5xum Mar 28 '17 at 13:11
  • But the definition given for a basis on wikipedia has nothing to do with topologies. The answer given by @user46944 makes sense to me. Do you disagree with what he says? – Pierre Mar 28 '17 at 13:15
  • 1
    @Pierre I agree with what user46944 is saying, yes. But the fact that the definition given for a basis on wikipedia has nothing to do with topologies is just plain false, since the first sentence on wikipedia says "In mathematics, a base (or basis) B for a *topological* space ...X" – 5xum Mar 28 '17 at 13:17
  • Ok so I hadn't read the introductory paragraph of the wikipedia page, I'd just skipped to the definition part. My understanding has changed. 'B is a base/basis for a topological space (X,T)' means that B is a subset of T such that every element of T can be written as a countable union of elements of B. It can then be shown that if B is a base for (X,T), then the two properties I originally stated in my question must hold. And furthermore, it can be shown that if both conditions hold, then there exists a unique T such that B is a base for (X,T). Am I now correct? – Pierre Mar 28 '17 at 13:36
  • @Pierre Yes, that's it! – 5xum Mar 28 '17 at 13:38
  • 1
    @Pierre "...such that every element...can be written as a countable union..." I think there's a mistake, as there's no need for the union to be countable. It can be any arbitrary union. – layman Mar 28 '17 at 16:20
2

There is an important distinction to be made, so please read my answer carefully.

Let $(X,T)$ be a topological space. Now, if we have a collection $\mathcal{B}$ of subsets of $X$ which satisfy properties $1$ and $2$ in your question, then these are called a "base to the set $X$", and they form their own topology, call it $T_{\mathcal{B}}$. This topology is the set of all possible unions of elements of $\mathcal{B}$.

Now, a natural question (which is one that you have) is: how is this new topology, $T_{\mathcal{B}}$, related to the original topology $T$?

Well, notice that when we defined $\mathcal{B}$ above, we didn't say all of the subsets in $\mathcal{B}$ have to be from $T$. If they are all from $T$, then since $T$ is closed under unions, it follows that $T_{\mathcal{B}} \subseteq T$. If in addition you want $T \subseteq T_{\mathcal{B}}$, you will need to show every element $A \in T$ can be written as a union of elements from $\mathcal{B}$, since this will imply $A \in T_{\mathcal{B}}$.

In general, though, you can have $T_{\mathcal{B}}$ be a different topology than $T$. You only get $T_{\mathcal{B}} \subseteq T$ if, as we said, the base $\mathcal{B}$ comes from sets in $T$. And you only get $T_{\mathcal{B}} = T$ if in addition to $\mathcal{B}$ coming from $T$, we also can write each element of $T$ as a union of elements of $\mathcal{B}$.

In general, $\mathcal{B}$ is always a basis for the topology $T_{\mathcal{B}}$ (the set of all possible unions of elements of $\mathcal{B}$) -- you should check that all possible unions of elements of $\mathcal{B}$ forms a topology. In the event that $T_{\mathcal{B}} = T$, we would then say $\mathcal{B}$ is a basis for the topology $T$.

layman
  • 20,191
  • Just to check I've understood: (1) Wikipedia defines a 'base B for a set X' (2) 'B is a basis for a topology T' means that T = the set of countable unions of elements from B – Pierre Mar 28 '17 at 12:59
  • If that is correct, I have one more question. Other than you, I can't seem to find a single source making a distinction between these two definitions. The phrases seem to be used interchangeably. Why is that? To me, that just adds to the confusion. Am I the only person who finds that confusing? – Pierre Mar 28 '17 at 13:04
  • @Pierre Sorry for not responding sooner, Pierre. It looks like you got your problem sorted by another person. For the record, you're not the only one confused by this. I was confused 2.5 years ago. See the question I asked: http://math.stackexchange.com/questions/923102/question-about-a-basis-for-a-topology-vs-the-topology-generated-by-a-basis – layman Mar 28 '17 at 16:18
0

The base $\mathcal{B}$ corresponds to the topology $\tau$ in the following way, for any $U\in \tau$, and for any $x\in U$, $\exists B\in \mathcal{B}$ such that $x\in B\subseteq U$. I hope this should clear the confusion.

Parish
  • 832