# Subbasis vs. basis

1. Feb 11, 2009

### tomboi03

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$$\cap$$B2.

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!

2. Feb 12, 2009

### phreak

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.

3. Feb 12, 2009

### Take_it_Easy

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.