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I am reading Topology by Munkres. The following lemma is proven:

A topology $\mathcal{T}$ equals all unions of elements of the basis $\mathcal{B}$.

The reasonign is like this: let $U \in \mathcal{T}$, choose for each $x \in U$ a basis element $B_x \in \mathcal{B}$. Then: $$U = \bigcup_{x \in U} B_x$$ I do not understand the final step. Shouldn't it be: $$U \subseteq \bigcup_{x \in U} B_x$$ Consider a topology on $X$ with a single basis element $B = X$. Then $U \subset B$, if $U \neq X$. Alternatively, pick an open subset of a basis: $U \subset B_i$. Then $U \subset \bigcup_{x \in U} B_x = B_i$.

Munkres defines a basis on $X$ as a collection $\mathcal{B}$ of subsets $B_i$ of $X$ satisfying: $$\forall x \in X \Rightarrow \exists B_i: x \in B_i$$ $$x \in B_1 \cap B_2 \Rightarrow \exists B_3: x \in B_3 \subset B_1 \cap B_2$$

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    $\begingroup$ What is the textbook's definition of basis? I think if $B=X$ is the only basis element, then the topology is just $\{X,\varnothing\}$, and the lemma would still be satisfied. $\endgroup$ – Matthew Leingang Mar 6 '17 at 19:47
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    $\begingroup$ Munkres says "$\mathcal{T}$ equals the collection of all unions of elements of $\mathcal{B}$." In quoting the proof, you also omitted the condition that $x \in B_x \subset U$, which explains the equality. One last thing, it's Munkres, not Munkers. $\endgroup$ – Fabio Somenzi Mar 6 '17 at 20:30
  • $\begingroup$ Seems to be a tautology, since a base for a topology $T$ is defined as any $B\subset T$ such that $T=\{\cup C: C\subset B\}.$ $\endgroup$ – DanielWainfleet Mar 7 '17 at 1:36
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A basis $\cal B$ for a particular topology $\cal T$ is a basis that further has the property that for each $U \in \cal T$ and for every $x \in U$ there exists $B \in \cal B$ with $x \in B \subset U$. Thus you would have $B_x \subset U$ for all $X$ and consequently $$\bigcup_{x \in U} B_x \subset U$$too.

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Munkres sort of has 2 notions of a base :

  1. We start with a topology $(X,\mathcal{T})$. A base for this particular space is a collection of open subsets $\mathcal{B}$ such that for each open set $U$ of $X$, and each $x \in U$, there exists some $B_x \in \mathcal{B}$ such that $x \in B_x \subset U$. One can then check that every $U \in \mathcal{T}$ is a union of base elements: for every $x \in U$ pick $B_x$ from the definition. Then $U = \cup \{B_x: x \in U\}$ because, as all $B_x \subset U$, we have the right to left inclusion and as each $x \in U$ is (At least) in its "own" $B_x$, we have the left to right inclusion. So a base is a more compact way to describe a topology. It also satisfies the 2 (necessary) conditions, because $X$ is open and the intersection of two base elements is open,so for each $x$ in it we have such a $B_3$.

  2. We can start with a set $X$. A collection of subsets $\mathcal{B}$ that satisfy those two conditions, is a base for some unique topology $\mathcal{T}$. It's clear how it should be defined: it's the collection of all unions of subcollections of $\mathcal{B}$. By definition $\mathcal{T}$ then contains $\emptyset$ (as the empty union), and is closed under all unions by construction, and the first condition gives that $X$ is in $\mathcal{T}$, while the second one implies that the intersection of two members of $\mathcal{T}$ is again in $\mathcal{T}$(some minor reasoning required), so this definition indeed gives a topology that $\mathcal{B}$ is a base for: any $O = \cup \mathcal{B'}$ for some $\mathcal{B'} \subset \mathcal{B}$. so any $X \in O$ is in some member of $\mathcal{B'}$ by definition. And it's clear by 1. that this is the only way we could have defined $\mathcal{T}$. Here we use a base to generate a topology ,not to describe a pre-given one more compactly.

Your question probably referred to the first notion: a base for a given space.

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