# Introduction to Topology- Part 2

So our last post familiarized us with the notion of a topology on a set. We know how to define a topology and a few basic examples of topological spaces. A natural problem arises when discussing topologies on large spaces. Suppose I asked you to list all open subsets of $\mathbb{R}$. This would take quite a while…forever, to be more precise. What we shall do instead, is provide a means of specifying the open sets from which all other open sets are derived. This basis allows us to describe fully the topology on $X$ by working only with a relatively small subset of the topology.

So what is a basis and how does is generate a topology?

Definition 1: If $X$ is a set, a basis, $\mathcal{B}$, is a collection of subsets of $X$ (elements of said collection are called basis elements) satisfying:

1. For each $x \in X$ there exists a $B \in \mathcal{B}$ so that $x \in B$.
2. Let $B_1,B_2 \in \mathcal{B}$. If $x \in B_1 \cap B_2$,  then there is another basis element, $B_{3}$ so that $x \in B_3$ and $B_{3} \subset B_{1} \cap B_{2}$.

We define the topology $\mathscr{T}$ generated by $\mathcal{B}$ to be the collection of all subsets of $X$, $U$, so that for each $x \in U$, there exists a $B \in \mathcal{B}$ so that $x \in B \subset U$. Thus, if this property holds, we call $U$ open (a member of the generated topology). Compare this to the usual definition of an open set in $\mathbb{R}$. That is, a set $U \subset \mathbb{R}$ is open, if, for each $x \in U$ there exists an $\epsilon >0$ so that $x \in B_\epsilon (x) \subset U$. Hmm…these look awfully similar. This should allude to a basis on $\mathbb{R}$, but we’ll touch on this in a moment.

It would be useful to in fact prove that the collection of all open sets as above is indeed a topology.

Theorem 1: The collection of all open subsets of a set $X$ described above forms a topology, $\mathscr{T}_\mathcal{B}$.

Proof: We proceed by showing that the axioms of a topology do indeed hold. Clearly $\emptyset \in \mathscr{T}_\mathcal{B}$ trivially. Also, $X \in \mathscr{T}_\mathcal{B}$, by definition of basis.

Next, we need see that arbitrary unions are indeed another element satisfying the above property. So, let $U = \bigcup \limits_{i \in I} U_i$. Then if $x \in U$, there exists an $i_0 \in I$ so that $x \in U_{i_0}$. Since $U_{i_0} \in \mathscr{T}_\mathcal{B}$, there exists $B \in \mathcal{B}$ so that $x \in B \subset U_{i_0} \subset U$. Ergo, $U \in \mathscr{T}_\mathcal{B}$.

Finally, we show that finite intersections are also in the topology. Let $U = \bigcap_{i=1}^{n} U_i$ and consider $x \in U$. Then, $x \in U_i$ for $i=1,\ldots,n$. Check for yourself that the second condition in the definition of a basis furnishes the following argument:

For each $i$ there exists $B_i$ so that $x \in B_i \subset U_i$. Condition two implies that there exists $B \in \mathcal{B}$ so that $x \in B \subset \bigcap_{i=1}^{n} B_i \subset U$. Thus, $U \in \mathscr{T}_\mathcal{B}$.

Now let’s look at a few examples of bases.

For the discrete topology on $X$, the basis is the collection of one-point sets in $X$. This basis property one trivially. Since our basis elements are all distinct points (and therefore disjoint sets), we know that the intersection of any basis gives the null set, so the third basis set in condition two is satisfied by the empty set.

Lets look at another example. Let $\mathcal{B}$ be the collection of all circular regions in the plane. Then $\mathcal{B}$ form a basis. Every point in the plane can be encompassed by a circle of certain radius from another point, satisfying property one. If a point belongs to the intersection of two arbitrary circles, then a smaller circle contained in the intersection around that point can be made. This satisfies property two of a basis, meaning that the set of circles in the plane form a basis. This topology will pop up again later when we study metric spaces.

So we have a few examples and the definition. Now, we have a means of describing two topologies. But how can we use this to compare them?

N.K/J.T.