Munkres states the following lemma in his Topology:

Lemma 13.2: Let $X$ be a topological space and suppose $K$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $K$ such that $x\in C\subset U$. Then $K$ is a basis for the topology of X.

I have shown that 1) Given $x \in X$ there is some open set in $K$ containing $x$.(2)Given $x\in C_1,C_2 \in K$, there is an open set $C \in K$ such that $x\in K \subset C_1 \cap C_2$.

Question: Shouldn't this be enough to prove that $K$ is a basis for the topological space $X$? After all, Munkres defines a basis as a collection of subsets of $X$ satisfying properties (1) and (2). Why do I need to show that that the topology generated by $K$ is the same as the given one?

  • $\begingroup$ In the book, the first paragraph of the proof shows that $\mathcal{C}$ is a basis. But a basis for $what$ space? A basis for some $Y$, $Z$ ? We must verify that it is a basis for $X$, and this is the content of the second paragraph f the proof. $\endgroup$ – Fei Li Aug 24 '16 at 10:58
  • $\begingroup$ Ok, I am a little confused now. In the definition (and I just noticed it), he says that if X is a set, a basis for a topology on X is a collection $\beta$ satisfying (1) Given $x \in X$ there is a $B \in \beta$ containing $x$.(2)Given $x\in C_1,C_2 \in \beta$, there exists a $C \in \beta$ such that $x\in C \subset C_1 \cap C_2$. So, I don't see how the basis needs a topological space. (It should but I don't see how the definition demands it) $\endgroup$ – user96343 Aug 24 '16 at 12:54
  • $\begingroup$ I've written up an answer. @user96343 $\endgroup$ – Fei Li Aug 24 '16 at 14:13

Read carefully the paragraph following the definition of basis on page$78$:

If $\mathcal{B}$ satisfies these two conditions, then we define the topology $\mathcal{T}$ generated by $\mathcal{B}$ as follows: A subset $U$ of $X$ is said to be ope in $X$ (that is, to be an element of $\mathcal{T}$) if for each $x\in U$, there is a basis element $B\in\mathcal{B}$ such that $x\in B$ and $B\subset U$. Note that each basis element is itself an element of $\mathcal{T}$.

In plain English, the whole definition and the paragraph says: Ok now, given the set $X$, let's define a topology on $X$ by a new trick: we specify which sets are open in $X$ indirectly in terms of basis. Once we have a basis on $X$ first, then we can define a topology $\mathcal{T}$ as follows: A subset $U$ of $X$ is said to be open in $X$ if……blah blah…...

So the definition has not predefined any topology on $X$. (The language of the definition is indeed prone to cause confusions. I remember I had struggle with it for a while when I was reading that as well) It just presents you a new trick to define topologies on $X$.

Now, in practice, often we already know some topology on $X$, and we would like to know:

Is there any basis that generates the topology we have?

So we want to go into the other direction. We would like to know that our old familiar topology indeed comes from this "new trick". Once we have verified that $\mathcal{C}$ is a basis, it automatically generates a topology $\mathcal{T}'$. But is it the product we want? Is it the same as our old $\mathcal{T}$? This is a nontrivial fact to be checked.

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  • $\begingroup$ Thank you very much! I think I understand it now. So, each time I see a question of the form "Is B a basis for top. space X?", I should read it as "Does it satisfy properties (1) and (2) and generate the same topology?". Right? $\endgroup$ – user96343 Aug 24 '16 at 16:21
  • $\begingroup$ Yes, exactly. This question is of practical importance because we might be able to find some $\mathcal{B}$ that has some nice properties, for example, being countable. Once we can prove that this nice $\mathcal{B}$ does indeed generate the already-defined topology, then we know that the space $X$ at hand is second countable (see page $190$), and this can greatly simplifies the proofs of the many desired properties we want for the space $X$. $\endgroup$ – Fei Li Aug 25 '16 at 9:08

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