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Topological and neighbourhood bases

  1. Jul 21, 2011 #1
    I'm trying to follow a proof in this video, #20 in the ThoughtSpaceZero topology series. I get the first part, but have a problem with second part, which begins at 8:16.

    Let there by a topological space [itex](X,T)[/itex]. Let [itex]x[/itex] denote an arbitrary element of [itex]X[/itex].

    Definition 1. Topological base. A set [itex]B \subseteq T[/itex] such that [itex](\forall T_i \in T) (\exists C \subseteq B) [T_i = \cup_C C_j][/itex].

    Definition 2. Neighbourhood base for [itex]x[/itex]. A subset [itex]\beta [x][/itex] of [itex]V[x][/itex], the neighbourhoods of [itex]x[/itex], such that [itex](\forall V_i \in V[x])(\exists B_i \in \beta [x])[B_i \subseteq V_i][/itex].

    Theorem. Let there be a topological space [itex](X,T)[/itex]. Let [itex]B \subseteq 2^X[/itex]. Let [itex]\beta [x] = \left \{ B_i \in B \;|\; x \in B_i \right \} \subseteq B[/itex]. Then [itex]B[/itex] is a topological base for T if and only if, for all [itex]x[/itex], the set [itex]\beta [x][/itex] is a neighbourhood base for [itex]x[/itex].

    Proof. Assume [itex]B[/itex] is a base for [itex]T[/itex]. [...]

    I understand that part; but I don't follow his proof of the converse. Paraphrasing here: (My comments in square brackets.)

    What if [itex]U_x \notin T[/itex]? Since [itex]U_x[/itex] is a neighbourhood of [itex]x[/itex], we could replace it with a subset of itself which is open, but how would we know that this subset of [itex]U_x[/itex] is in [itex]B[/itex]?
    Last edited: Jul 21, 2011
  2. jcsd
  3. Jul 21, 2011 #2
    What he meant to say is that [itex]\beta(x)\subseteq B[/itex]. Calling it a base is wrong (since that's what we wanted to prove).

    Well noticed!! Indeed, one should take B a subset of T from the very beginning. That is, the theorem should be

    Let [itex](X,\mathcal{T})[/itex] be a topological space and let [itex]\mathcal{B}\subseteq \mathcal{T}[/itex]. Then [itex]\mathcal{B}[/itex] is a base if and only if
    [tex]\beta (x)=\{B\in \mathcal{B}~\vert~x\in B\}[/tex]
    is a neighbourhood base for all x
  4. Jul 21, 2011 #3


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    I wrote this before micromass replied.

    I don't have time to study the proof in detail, but I have some notes on this that I can (almost) just copy and paste, so I'll do that. Maybe it will help, maybe it won't.

    Theorem: Suppose that [itex](X,\tau)[/itex] is a topological space. The following conditions on a set [itex]\mathcal B\subset\mathcal P(X)[/itex] are equivalent.

    (a) Every non-empty open set is a union of members of [itex]\mathcal B[/itex].
    (b) If [itex]E[/itex] is open and [itex]x\in E[/itex], there's a [itex]B\in\mathcal B[/itex] such that [itex]x\in B\subset E[/itex].

    (a) [itex]\Rightarrow[/itex] (b): Let [itex] E[/itex] be an arbitrary non-empty open set, and let [itex]x\in E[/itex] be arbitrary. By assumption, there's a set [itex]\{B_i\in\mathcal B|i\in I\}[/itex] such that [itex]E=\bigcup_{i\in I} B_i[/itex]. This means that there's a [itex]j\in I[/itex] such that [itex]x\in B_j\subset\bigcup_{i\in I} B_i=E[/itex].

    (b) [itex]\Rightarrow[/itex] (a): Let [itex]E[/itex] be an arbitrary non-empty open set. For each [itex]x\in E[/itex], choose [itex]B_x\in B[/itex] such that [itex]x\in B_x\subset E[/itex]. The fact that [itex]x\in B_x[/itex] for all [itex]x\in E[/itex] implies that [itex]E\subset\bigcup_{x\in E} B_x[/itex]. The fact that [itex]B_x\subset E[/itex] for all [itex]x\in E[/itex] implies that [itex]\bigcup_{x\in E} B_x\subset E[/itex]. So [itex]\bigcup_{x\in E} B_x=E[/itex].

    Definition: If [itex](X,\tau)[/itex] is a topological space, a set [itex]\mathcal B\subset\tau[/itex] that satisfies the equivalent conditions of the theorem is said to be a base for the topology [itex]\tau[/itex].

    I wasn't even aware that there's a term for a collection of subsets that satisfies (a) and another one for a collection of subsets that satisfies (b).

    Edit: I fixed a mistake in the statement of the theorem. I had left out the word "non-empty" from condition (a). It should definitely be there.
    Last edited: Jul 21, 2011
  5. Jul 21, 2011 #4
    That is not a problem because of some trickery. In fact, (a) doesn't imply that [itex]\emptyset\in \mathcal{B}[/itex]. We can always write

    [tex]\emptyset = \bigcup \{B\in \mathcal{B}~\vert~B\neq B\}[/tex]

    The statement [itex]B\neq B[/itex] isn't really important, the importance is to take an empty union! The fun thing is that the empty union is always empty. So if we demand that every open set is a union of sets, then this is always true for the empty set!!
  6. Jul 21, 2011 #5


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    That's a fun trick. I certainly didn't think of that. :smile: But I think I prefer to just add the word "non-empty" to (a).
  7. Jul 21, 2011 #6
    Thanks all for clearing that up!

    (I've just made some edits of my own to the OP. Must have been too intent on getting the LaTeX right to notice that I wrote "neighbourhood basis for x if and only if [...] is a neighbourhood basis for x". Yikes!)
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