# Are basis elements of a topology always open?

I'm currently going through Munkres' Topology, and his discussion of bases for topologies has me a little confused. He defines a basis $\mathfrak{B}$ for a topology $\mathscr{T}$ on a set $X$ as a collection of subsets that satisfy the given requirements. These elements of $\mathfrak{B}$ are not defined as open, but in the following lemma which claims that $\mathscr{T}$ equals the collection of all unions of elements from $\mathfrak{B}$, he claims in the proof of this lemma that the elements of $\mathfrak{B}$ are in $\mathscr{T}$, and therefore all of the basis elements are open.

Why is this true? I know the basis elements are open in the topology generated by $\mathfrak{B}$, but the lemma refers to an arbitrary topology on some set X. Why should I assume that these are open sets? I appreciate any help. Thanks.

• The given requirements are the definition of basis elements. I didn't want to write the entire definition out. Munkres only claims that $\mathfrak{B}$ is a subset of the set X, not the topology $\mathscr{T}$, and he does not refer to X as a topological space. – Newman Apr 23 '17 at 22:10
• He defines a basis (for some topology), not a basis for a (given) topology, which is another matter. That is, he defines a basis, and then the topology comes later. You could start with a topology, and define a basis for that topology, but this is a different definition. – Mirko Apr 23 '17 at 22:14
• Right, but this doesn't seem like enough to claim that the basis elements are open, as Munkres does in the proof of the lemma. – Newman Apr 23 '17 at 22:16
• As shown in the answer by @Sheaf, a basic element $B$ is open, since $B\subseteq B$. – Mirko Apr 23 '17 at 22:17
• There is a viewpoint that basis elements need not be open, this is how Bourbaki view it: math.stackexchange.com/questions/80025/…. – Moishe Kohan Apr 24 '17 at 19:44

Here is your confusion: what Munkres starts out with is a set $X$ and a collection $\mathfrak B$ of subsets of $X$, which satisfy the given properties. Only then do we define the topology $\mathscr T$ using this basis, so until this is done it doesn't make sense to talk about the basis elements of $\mathfrak B$ as being "open".

Now, once we do define our topology, recall that we define a subset $U$ of $X$ to be open if given $x\in U$ there exists $B\in\frak B$ such that $x\in B\subseteq U$. Since this certainly holds when $U=B\in\frak B$, we see that all elements of $\frak B$ are open.

This is the precise statement of the lemma you are referring to:

Let $X$ be a set; let $\frak B$ be a basis for a topology $\mathscr T$ on $X$. Then $\mathscr T$ equals the collection of all unions of elements of $\frak B$.

So when Munkres says this, he is doing as we say above: that is, he is starting with some collection of subsets $\frak B$ that generate a topology, and he is calling this topology $\scr T$. So by what we said above, all basis elements are open in $\scr T$ and there should be no issue.

• Right, the basis elements are open in the topology generated by $\mathfrak{B}$. However, in the lemma he proves following the definition it is not clear that the topology referenced is the topology generated by $\mathfrak{B}$. – Newman Apr 23 '17 at 22:14
• @Newman see my edit, and let me know if I am still misunderstanding your question. – Alex Mathers Apr 23 '17 at 22:16
• OK, so it is implied that this topology is in fact the topology generated by $\mathfrak{B}$, rather than an arbitrary topological space? – Newman Apr 23 '17 at 22:17
• @Newman Yes that's correct. If you move on to the next lemma, you'll see a situation where we start out with a topology $\scr T$ in advance, and we are curious when a collection of subsets indeed forms a basis that will generate $\scr T$ – Alex Mathers Apr 23 '17 at 22:18
• @Newman I guess the key thing that may be confusing here, is that Munkres is using "$\frak B$ generates $\scr T$" and "$\frak B$ is a basis for $\scr T$" interchangeably. – Alex Mathers Apr 23 '17 at 22:20