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I'm currently looking at Lemma 13.2 in Munkres' Topology. It states the following: Given a collection $C$ of open sets of a topological space $X$ such that for each open set $U$ of $X$ and each $x$ in U, there is an element $C'$ of $C$ such that x $\in C' \subset U$. Then $C$ is a basis for the topology of $X$.

In the proof, it is both shown that $C$ is a basis and that the topology generated by $C$ is equal to the collection of open sets of X. What is the purpose of the second part of the proof? Is it because a particular topology ("the" topology) is specified for X in the lemma? I'm confused because the definition of a basis for a topology on X doesn't mention this.

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You start with a topology $\tau$ on $X$. The basis $C'$ generates a new (a priori different) topology $\tau'$. The question is whether $\tau = \tau'$, i.e., whether $C'$ is a basis for the original topology $\tau$. –  Martin Jan 6 '13 at 8:09
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Suppose you are starting with the space $(X, \tau)$. You have to prove two things: $\mathcal{C}$ is a base for some topology on $X$: Munkres has a theorem for this, with conditions to check that only depend on the family $\mathcal{C}$ in question, and it generates the topology $\tau$ that we started with. This is what the statement "$\mathcal{C}$ is a base for $(X, \tau)$" means.

Supposing we have checked that the collection $\mathcal{C}$ satisfies the (intrinsic) conditions for being a base for some topology, it is clear that the topology $\mathcal{C}$ generates, say $\tau'$, is a subset of $\tau$, as we start with subsets that are in $\tau$, so the smallest topology that contains $\mathcal{C}$ (which we called $\tau'$) is a subset of $\tau$ (as the latter is one of the candidates).

On the other hand, the condition on $\mathcal{C}$ can just be reformulated as "every open set of $\tau$ is a union of members from $\mathcal{C}$", which easily shows that any topology (closed under unions!) that contains $\mathcal{C}$ also contains all members of $\tau$, showing the other conclusion.

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Does the converse of the theorem hold true, that if $C$ is a basis for a topology on $X$,then for each open set $U$ of $X$ and each $x \in U$, there is an element $C′$ of $C$ such that $x \in C′⊂U$. I am not sure of this, because this looks like a much more stronger condition for a basis, then that defined previously, but Munkres seems to have used this in theorem 15.1. Thanks. –  ramanujan_dirac Jan 28 '13 at 11:29
    
@ramanujan_dirac Indeed $\mathcal{C}$ is a base for the topology $\mathcal{T}$ iff for every set $O \in \mathcal{T}$ and for every $x \in O$ there is some $C \in \mathcal{C}$ with $x \in C \subset O$. This holds because we have to be able to write every element of $\mathcal{T}$ as a union of members of $\mathcal{C}$, and this condition just expresses that. The Munkres condition is for a collection of sets to be a base for some (to be defined) topology, not a pre-given one. –  Henno Brandsma Jan 28 '13 at 19:34
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Munkres first defines what is meant by a basis for a topology on a set $X$. That is, starting with a set $X$ and such a collection $\mathcal{B}$ of subsets of $X$ one can unambiguously define the topology on $X$ generated by $\mathcal{B}$ (by taking unions of its subfamilies). Of course, sometimes topologies are already given to you, and it might be interesting to know how they can be generated. This leads us to the following

Question: Suppose we have a topology $\mathcal{T}$ on a set $X$. Which bases for topologies on $X$ actually generate the same topology $\mathcal{T}$?

Lemma 13.2 provides a way of determining the answer to this question. Once one has answered this question, one may also consider the following:

Question: Given a topology $\mathcal{T}$ on a set $X$, what is the smallest cardinality of a basis for for $\mathcal{T}$?

It turns out that an answer to this question often leads to a much greater understanding of some properties of the space in question without having to delve into specifics, especially if the topology has a countable basis.

As an example, consider the $\mathbb{R}$ with the usual (metric) topology.

  • Note that the collection of all open intervals centred at $0$ (i.e., $\mathcal{B} = \{ ( -a , a ) : a > 0 \}$) forms a basis for a topology on $\mathbb{R}$. However this collection clearly does not generate the usual topology on $\mathbb{R}$. (The set $(0,1)$ is not open in this topology.)

  • Note that the collection of all half-open intervals $[a,b)$ ($a < b$) also forms a basis for a topology on $\mathbb{R}$. Again, the topology generated by this basis is not the usual topology (it is a finer topology called the lower limit (or Sorgenfrey) topology.)

  • Clearly the collection of all (metric) open subsets of $\mathbb{R}$ forms a basis for a topology on $\mathbb{R}$, and the topology generated by this basis is the usual one. However this is somewhat unsatisfying because we have not really simplified our description of the topology. (Part of the reason we sometimes describe a topology by just listing a basis is because sometimes it is troublesome to exactly describe what we mean by an open set; for example, properties that open sets must have might depend on the points that they contain.)

  • The family of all open intervals in $\mathbb{R}$ with rational endpoints is also a basis for a topology on $\mathbb{R}$, and it generates the usual topology. An important thing to note is that this basis is countable which actually tells us quite a bit about the topological space: given $A \subseteq \mathbb{R}$, if $x \in \overline{A}$, then there is a sequence in $A$ which converges to $x$; there is a countable dense subset of $\mathbb{R}$; there are (at most) $2^{\aleph_0}$ open subsets of $\mathbb{R}$, ... (Of course, you have likely seen these particular properties of the real line earlier, but any topological space with a countable basis shares these properties.)

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Generated is too general for base. Generated by just taking unions is more correct, I think. Otherwise we allow subbases as well, and these are usually considered separately. –  Henno Brandsma Jan 6 '13 at 8:48
    
@Henno: You're probably right. I've made a slight alteration. Thanks. –  Arthur Fischer Jan 6 '13 at 9:15
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