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Topology proofs

  1. Aug 2, 2008 #1
    i've texed up three proofs in from elementary topology. can someone please check them?

    actually i'll just retype them here for convenience

    8.2.5

    Let [itex] f: X_{\tau} \rightarrow Y_{\nu} [/itex] be continuous and injective. Also let [itex] Y_{\nu} [/itex] be Hausdorff.

    Prove : [itex] X_{\tau}[/itex] is Hausdorff.

    Proof : Pick [itex]y_1[/itex] and [itex] y_2 [/itex] in [itex] f(X) [/itex]. By injectivity of f there exist [itex]x_1 =[/itex]
    [itex]f^{-1}(y_1)[/itex] and [itex] x_2 = f^{-1}(y_2) [/itex] such that they are both unique. [itex] f(X) [/itex] is Hausdorff so by Theorem 1 there exist disjoint open neighborhoods [itex] U [/itex] and [itex] V [/itex] of [itex]y_1[/itex] and [itex] y_2 [/itex] respectively. Then [itex] f^{-1}(U \bigcap V) = f^{-1}(U) \bigcap f^{-1}(V) = \emptyset [/itex] and by continuity [itex] f^{-1}(U) [/itex] and [itex] f^{-1}(V) [/itex] are open. Finally by definition of [itex] f^{-1} : x_1 \in f^{-1}(U) [/itex] and [itex] x_2 \in f^{-1}(V) [/itex] which, as stated previously, are two open disjoint sets in [itex] X_{\tau} [/itex]. Hence [itex] X_{\tau} [/itex] is Hausdorff.\\



    Comments:

    Injectivity is necessary. Take for example [itex] X = \{a,b\} [/itex] and [itex] Y = \{a\} [/itex] and to be [itex] f: X_{\mathcal{I}} \rightarrow Y_{\mathcal{I}} [/itex]. Explicitly [itex] f(\{a,b\}) =\{a}\}[/itex]. f is continuous, [itex] Y_{\mathcal{I}} [/itex] is obviously Hausdorff and [itex] X_{\mathcal{I}} [/itex] is obviously not.\\

    This does not prove that Hausdorff is a strong topological property because we have proven a stronger converse. To prove that Hausdorff is a strong topological property we would have to have proven that [itex] f: X_{\tau} \rightarrow Y_{\nu} [/itex] continuous, not necessarily injective, and [itex] X_{\tau}[/itex] Hausdorff implies [itex] f(X) [/itex] Hausdorff.

    8.2.7


    Let [itex] X_{\tau}[/itex] be [itex] T_1 [/itex] and [itex] A \subseteq T [/itex] and [itex] x \in A' [/itex].\\

    Prove : Any neighborhood of x intersects \textit{A} in infinitely many points.

    Proof : Assume that there exists a neighborhood of x, in [itex] X_{\tau}[/itex] that intersects \textit{A} in only finitely many points to derive a contradiction. Let [itex] N_x [/itex] be such a neighborhood. [itex] N_x [/itex] is [itex] T_1 [/itex] by Theorem 1. Hence we can seperate x from all points in [itex]N_x [/itex] by other neighborhoods. Since there are finitely many points in [itex] N_x [/itex] there are finitely many such neighborhoods. Let \textit{N} be the intersection of those neighborhoods. [itex] N -\{x\}[/itex] therefore is itself a non-trivial open neighborhood of \textit{x} which does not intersect \textit{A}. This contradicts that x is a limit point of \textit{A} and therefore any neighborhood of x intersects \textit{A} in infinitely many points.



    \textbf{8.3.4}\\

    Let [itex] A \subseteq X_{\tau}[/itex] and [itex]X_{\tau}[/itex] be regular.\\

    Prove : A is regular.

    Proof : Pick [itex] A_1 \subseteq A [/itex], [itex]A_1[/itex] closed in the subspace topology, and [itex] x \in A - A_1 [/itex]. Then [itex]A_1 = B[/itex] for some [itex]B \in \tau[/itex] and we can find two open sets [itex]N_B[/itex] and [itex]N_x[/itex] by the the regularity of [itex]X_{\tau}[/itex] which are disjoint. [itex]N_B \bigcap A[/itex] and [itex]N_x \bigcap A[/itex] are two disjoint sets in A which contain [itex]A_1[/itex] and x respectively. Therefore A is regular.


    Theorem 1 : A subspace of a [itex]T_i[/itex] space for [itex]i \leq 2[/itex] is [itex]T_i[/itex].
     

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  3. Aug 3, 2008 #2
    I'm not an expert at all... but in 8.2.5 shouldn't you say "pick [tex]y_{1} \neq y_{2}[/tex] in [tex]f(X)[/tex]?
     
  4. Aug 3, 2008 #3
    yea you're right
     
  5. Aug 3, 2008 #4
    When you want to prove that X is hausdorff I think it is nicer to say: Pick [itex] x_1 \neq x_2[/itex] in X, and then do as you do. It is because when it is X you want to prove is hausdorf I think you should start there, because you have to prove it for any two x'es in X.
     
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