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Lemma 13.2 from Munkres

  1. Aug 1, 2011 #1
    Lemma 13.2: Let X be a topological space. Suppose that C 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 C such that [itex]x\in c\subset U[/itex]. Then C is a basis for the topology of X.

    Proof: The first paragraph is trivial, it just shows that the conditions of basis are satisfied.

    The second paragraph attempts to show that [itex]\tau'[/itex], the topology generated by C, is the same as the topology [itex]\tau[/itex] on X.

    Can someone elaborate on the second paragraph please!?!
  2. jcsd
  3. Aug 1, 2011 #2
    What don't you understand about the second paragraph??

    You just let [itex]\mathcal{T}^\prime[/itex] be the topology generated by the basis [itex]\mathcal{C}[/itex]. So by definition, a set G belongs to [itex]\mathcal{T}^\prime[/itex] if for every x in G, there is a set C in [itex]\mathcal{C}[/itex] such that

    [tex]x\in C\subseteq G[/tex]

    So, to prove that [itex]\mathcal{T}\subseteq\mathcal{T}^\prime[/itex]. Take G in [itex]\mathcal{T}[/itex]. By hypothesis, there exists for every for every x in G, a set C in [itex]\mathcal{C}[/itex] such that

    [tex]x\in C\subseteq G[/tex]

    So, by definition almost, we have that [itex]\mathcal{T}\subseteq \mathcal{T}^\prime[/itex].

    The other inclusion is less obvious. So to prove that [itex]\mathcal{T}^\prime\subseteq\mathcal{T}[/itex] we take an element [itex]G\in \mathcal{T}^\prime[/itex]. By hypothesis, there is for every x in G, a set C in [itex]\mathcal{C}[/itex] such that

    [tex]x\in C\subseteq G[/tex]

    This implies that [itex]G=\bigcup{C}[/itex] is the union of all these C's. Since all the C's are in [itex]\mathcal{T}[/itex], it implies that G is also in [itex]\mathcal{T}[/itex] as union of open sets...

    IS that more clear?
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