Difference Between Subbasis and Basis in Topology

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This is probably a stupid question.. but

Can someone tell me the difference between subbasis and basis.. in topology?? I know the definitions...

So Subbasis is defined to be the collection T of all unions of finite intersections of elements of S (subbasis)

sooo... S is pretty much a topology on X which is a collection of subsets of X whose union equals X.

Basis, however... is
If X is a set, basis on X is a collection B of subsets of X (basis elements) s.t.
1. for each x \in X, there is at least one basis element B containing x.
2. If x belongs to the intersection of two basis elements B1 and B2, then there is a basis element B3 containing x such that B3\subset B1\capB2.

Right? So pretty much... A subset U of X is said to be open in X if for each x \in U, there is a basis element B \in B such that x \in B and B \subset U.

But I'm still not understanding this quite... so well..

Can someone explain this to me??


Thank You!
 
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The topology generated by a subbasis is the topology generated by the basis of all finite intersections of subbasis elements. It seems like you just need an example, though, so here's one.

A basis for the standard topology of R is the collection of all open intervals. A subbasis for R is the collection \{(a,\infty):a\in \mathbb{R}\} \cup \{(-\infty,b):b\in \mathbb{R}\}. The reason that this is a subbasis for R is because finite intersections of elements in this set are precisely the basis elements of R. For instance, we have (a,b) = (-\infty,b)\cap (a,\infty). Subbases are often important because they offer an easier way of expressing a topology. In the previous example, there are many fewer sets in the sub-basis than there are in the basis.
 
I think there is a distiction that can help you avoid confusion.

If you have a set X and NO topology on it
every S subset of the set P(X) of the parts of X
can be taken and you can generate e topology that has S as subbasis.

If you have a set X and NO topology on it then
if you have a set B \subset P(X)
that satisfies the 2 property 1. ans 2. you mentioned you can generate a topology on
X that has B has basis.


Now if you have a set X and HAVE a topology T on it
a set S \subset P(X) is a subbasis of (X,T) if and only if (by definition)
for every open set A and for every a \in A exists
S_1, \ldots, S_n \in S such that
a \in S_1 \cap \cdots S_n \subset A.


Now if you have a set X and HAVE a topology T on it
a set B \subset P(X) is a basis of (X,T) if and only if (by definition)
for every open set A and for every a \in A exists
B_1 \in B such that
a \in B_1 \subset A.



There is a slight difference to understand.
Please read it carefully and think on it and you'll get the concept.
Hope this helps.
 
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