I have some questions regarding bases in topology. There are different ways of defining a topology and I want to distinguish between them.

Let $X$ be a topological space. A collection $\mathcal{B}\subseteq 2^{X}$ is called a basis for the topology on X if the following are satisfied:

1) Every element of $\mathcal{B}$ is an open subset of $X$.

2) Every open subset of $X$ is the union of some collection of elements of $\mathcal{B}$.

Let $X$ be a set, and suppose $\mathcal{B}\subseteq 2^{X}$. Then $\mathcal{B}$ is a basis for some topology on $X$ iff it satisfies the following:

1) $\bigcup\limits_{B\in\mathcal{B}}B=X$

2) If $B_1,B_2\in\mathcal{B}$ and $x\in B_1\cap B_2$, then there exists an element $B_3\in\mathcal{B}$ such that $x\in B_3 \subseteq B_1\cap B_2$.

When asked to prove that a set is a basis for the topology on a top. space $X$, we should use the first definition. However, in this case, do we need to also show the "basis criterion?"

(Basis Criterion) Suppose $X$ is a topological space, and $\mathcal{B}$ is a basis for its topology. Show that a subset $U\subseteq X$ is open iff it satisfies the following condition: For every $x\in U$, there exists $B\in\mathcal{B}$ such that $x\in B\subseteq U$.

When should we used the second definition (proposition)?

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    $\begingroup$ The second is mainly used to create a topology when you don't already have one. $\endgroup$ – Randall Oct 22 '17 at 1:11

That proposition will only prove that your set $\mathcal{B}$ is the basis for some topology. This doesn't mean it will actually be a basis for the topology you are in fact considering.

The basis criterion along with all sets in $\mathcal{B}$ being open is equivalent to your first definition. So you have the following proposition which is easy to prove.

Let $(X, \tau)$ be a topological space and a set $\mathcal{B} \subset \tau$. Then $\mathcal{B}$ is a basis for the topology $\tau$ on $X$ if and only if for every open $U$ and every $x \in U$, there exists a $B \in \mathcal{B}$ such that $x \in B \subset U$.

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    $\begingroup$ I have a question in this context, suppose we have two topologies with same base, then will we conclude that the two topologies equal, as we know that the base determines an unique Topology ? @Demophilus $\endgroup$ – hiren_garai Oct 22 '17 at 1:43
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    $\begingroup$ Yes that is true, there is even the following, slightly more general result: Let $\mathcal{B}_1$ and $\mathcal{B}_2$ be two bases that generate the topologies $\tau_1$ and $\tau_2$ respectively. If $\mathcal{B}_1 \subset \mathcal{B}_2$ then $\tau_1 \subset \tau_2$. $\endgroup$ – Demophilus Oct 22 '17 at 9:20
  • $\begingroup$ Okay ! Thank you for your answer !. $\endgroup$ – hiren_garai Oct 22 '17 at 10:36

If you're given a topological space, meaning you're given the topology, you certainly need to use the first definition, as you state.

You should use the second statement if you are given a set $X$ but not told its topology. I can imagine a question like:

"Here is a set $X$, and here is a subset of $2^X$. Is this subset of $2^X$ a basis for a topology?"

Here you can't use the first definition at all, because you don't have any notion of what the open subsets are. You're being asked "If I give you this basis, can you give me a topology?" as opposed to "Here's a topology, is this a basis for the topology I have given you?".

Your "Basis criterion" appears to be talking about sets $U$ (which you haven't defined - I'm guessing these are your open sets in the topological space $X$?), which means you already have a topology, and so you are not in the situation to use your second definition. You need this criterion to hold in order that your thing is a basis for the topology you've been given.


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