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Is a convex subset of a connected space connected?

  1. Jan 31, 2012 #1
    It seems like something that could (should?) be true, but with topology you never know (unless you prove it...).

    EDIT: I'll be more exact: let [itex](X,\mathcal T)[/itex] be a topological space with X a totally ordered set and [itex]\mathcal T[/itex] the order topology. Say X is connected and [itex]A \subset X[/itex] is convex (i.e. [itex]\forall a,b \in A: a < b \Rightarrow [a,b] \subset A[/itex]). Is A connected?
     
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  3. Jan 31, 2012 #2

    micromass

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    Let X be an ordered set equiped with the order topology. If X is connected, then you can show that

    1) Every set that is bounded from above has a least upper bound
    2) If x<y, then there is a z such that x<z<y.

    Try to prove this.

    Then you can prove that all intervals are connected just as you prove this with [itex]\mathbb{R}[/itex].
     
  4. Jan 31, 2012 #3
    For future reference:

    1) Let A be a subset of X bounded from above but without least upper bound, then define the non-trivial subset B as the set of all upper bounds.
    B is closed: take a [itex]b \in \overline B[/itex], and suppose there is an [itex]a \in A[/itex] with b < a, then [itex]]- \infty, a[[/itex] is an open around b that does not intersect B. Contradiction, hence b is an upper bound of A.
    B is open: take a [itex]b\in B[/itex], since b is not the smallest upper bound, there is a [itex]\beta \in B[/itex] with [itex]\beta < b[/itex], hence [itex]b \in ]\beta,+ \infty[ \subset B[/itex].
    Contradiction.

    2) Quite obvious indeed :) otherwise one can construct a separation of X.

    By golly you're right. As you probably know, the converse is a theorem in Munkres (section 24), but I hadn't realized it was a characterization. So every connected order topology corresponds to a linear continuum. Nice to know! Thanks :)
     
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