Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rao: Proposition 1.2.4. Superfluous section of proof?

  1. Jun 19, 2012 #1
    Rao: Topology: Proposition 1.2.4. If (X,T) is a topological space, a subset A of X is closed iff the the derived set of A is a subset of A: [itex]A'\subseteq A[/itex].

    Rao's proof of [itex](A'\subseteq A) \Rightarrow (X\setminus A \in T)[/itex] goes like this:

    To me, this looks like enough to show that [itex](A'\subseteq A) \Rightarrow (X\setminus A \in T)[/itex], since a set A is open iff each of its points belongs to a neighborhood which is a subset of A. So [itex]X\setminus A[/itex] is open. In other words, A is closed.[/QUOTE]

    But Rao goes on:

    This seems superfluous to me. Am I missing something? Why not just say U is the neighborhood of x that's a subset of [itex]X\setminus A[/itex]?
     
  2. jcsd
  3. Jun 19, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Your proof looks fine. What Rao says isn't wrong, but it can be shortened.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rao: Proposition 1.2.4. Superfluous section of proof?
  1. Conic sections (Replies: 4)

Loading...