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I've looked on this site but no answer seems to clarify my doubt or maybe I need someone to dumb it down a little. What I understand is that a basis is the collection of open sets whose union generates the topology while the subbasis is the collection of sets whose union of finite intersections generates the topology. Now I don't know how much to believe that because according to me the basis need not contain the entire set on which the topology is described while I did find such a description somewhere. I then decided to illustrate it for myself to try and understand but here too I got stuck:

Let $X$ be a set on which a topology $\mathcal{T}$ is defined $\mathcal{T} = \{ \varnothing, \{1,2\}, \{2,3\}, \{2\}, \{1,2,3\}, \{1,2,3,4\} \}$. In this case is $\mathcal{B} = \{ \{1,2\}, \{2,3\}, \{2\}, \{1,2,3\}, \{4\} \}$? Because if so $X$ doesn't belong to it which is what I read however according to this Wikipedia quote it should

The collection of open sets consisting of all finite intersections of elements of subbasis $\mathcal{S}$ together with the set $X$, forms a basis for $\mathcal{T}$.

But I don't see any reason why $X$ should belong to it. Further what would the subbasis look like? According to the above definition I should have some element in $\mathcal{S}$ whose intersection with another element gives the element $\{1,2,3\}$ in $\mathcal{B}$ but I see not two elements that can produce this intersection. Please help.

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B is a base of topology $\;T\;$ on a space $\;X\;$ = any open set in $\;X\;$ ( i.e., any element of $\;T\;$) can be expressed as a union of element in $\;B\;$ .

S is a sub base of topology $\;T\;$ on a space $\;X\;$ = any open set in $\;T\;$ can be a expressed as a union of finite intersections of elements in $\;S\;$ .

Thus for example, $\;B:=\{\;(a,b)\;/\;a,b\in\Bbb R\;\}\;$ is a base for the usual, Euclidean topology of the real line $\;\Bbb R\;$ , and $\;S:=\{\;(-\infty,a)\,,\,(b,\infty)\;/\;a,b\in\Bbb R\;\}\;$ is a sub base for the same topology.

Yet this is one concept that can vary from author to author...perhaps because it is not that useful (and thus it is not that important) in general...

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I suspect they use the convention that the intersection of the empty family equals the whole space, i.e.:

$$\bigcap \{\} = \bigcap \emptyset = X$$

This makes sense because of the same reason that the infinum of the emptyset is the greatest element of a set.

If this convention is not made, you are right that this is wrong. The construction to obtain a subbasis would rather be to add the emptyset and the whole space with the set you begin with.

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  • $\begingroup$ I'm not familiar with this convention. $\endgroup$ – user43470 Mar 14 '18 at 9:14
  • $\begingroup$ @user43470 Then this is a nice opportunity to get familiar with it. $\endgroup$ – drhab Mar 14 '18 at 9:15
  • $\begingroup$ Now you are. This is a generally accepted convention that makes life a lot more pleasant in topology (don't worry about it too much, I started to learn topology a month ago and I found it weird first too) $\endgroup$ – user370967 Mar 14 '18 at 9:16
  • $\begingroup$ Your suspicion is correct. $\endgroup$ – drhab Mar 14 '18 at 9:17
  • $\begingroup$ I find it both hard to understand and believe. So are we saying that we're taking the intersection of null with itself? I really didn't understand that proof on top. How can x belong to y? $\endgroup$ – user43470 Mar 14 '18 at 9:21
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An empty intersection is also a finite intersection.

If we are "working" in some space $X$ then we apply the convention: $$\cap\varnothing=X$$

Is it true that every element $x\in X$ is an element of $\cap\varnothing:=\{x\mid\forall y\in\varnothing[x\in y]\}$?

Yes, because no element $y\in\varnothing$ can be found with $x\notin y$, simply because $\varnothing$ has no elements.

Is it true that every element of $\cap\varnothing$ is a an element of $X$?

Mmm..., what exactly is $\cap\varnothing$? Actually it can be shown that every set belongs to it, but the set of sets does not exist so..

Here the convention comes in: let us agree that - if we are working in some universe - $\cap\varnothing$ stands for that universe.

In this context this leads to the agreement that $X$ also is an element of the collection of finite intersections of elements of some subbase.


edit (to give some perspective on subbase/base of topology):

If you start with a set $X$ and some collection $\mathcal V\subseteq\wp(X)$ then there is smallest topology on $X$ that contains $\mathcal V$.

Then $\mathcal V$ is a subbase for that topology.

It can be proved that the topology consists of unions of finite intersections of elements of $\mathcal V$, and the topology is said to be generated by $\mathcal V$.

According to the convention described above $X$ will belong to this collection.

Now if $\mathcal V$ is closed under binary intersections and secondly satisfies $\cup\mathcal V=X$ then every finite intersection of elements of $\mathcal V$ can be written as a union of elements of $\mathcal V$.

Consequently the collection of unions of finite intersections of elements of $\mathcal V$ coincides with the collection of unions of elements of $\mathcal V$.

In that special case $\mathcal V$ serves not only as subbase of the generated topology, but also serves as base of that topology.


If on the other hand you start with a topology on $X$ then the question rises: what collections would generate this topology?

Every collection that does is by definition a subbase of that topology.

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  • $\begingroup$ By smallest topology do you mean like the coarsest? Could you please give an example? $\endgroup$ – user43470 Mar 14 '18 at 10:03
  • $\begingroup$ Yes, the coarsest. If e.g. $X=\mathbb R$ and $\mathcal V=\{\{1,2\},\{2,3\}\}\subseteq\wp(\mathbb R)$ then the collection of finite intersections is: $\mathcal B:=\{\{2\},\{1,2\},\{2,3\},\mathbb R\}$. This collection automatically serves as a base in the sense that it has the properties described here. The topology generated is then $\{\varnothing,\{2\},\{1,2\},\{2,3\},\{1,2,3\},\mathbb R\}$. Its elements are unions of elements of base $\mathcal B$. $\endgroup$ – drhab Mar 14 '18 at 10:52

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