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

Invariance of domain

  1. Mar 19, 2012 #1
    In the first volume of Differential Geometry, Ch. 1, Spivak states that if [itex]U \subset \mathbb{R}^n[/itex] is homeomorphic to [itex]\mathbb{R}^n[/itex], then [itex]U[/itex] is open. This seems obvious: [itex]\mathbb{R}^n[/itex] is open in [itex]\mathbb{R}^n[/itex], so its pre-image under a homeomorphism [itex]f:U \rightarrow \mathbb{R}^n[/itex] is open. The pre-image under [itex]f[/itex] of [itex]\mathbb{R}^n[/itex] is [itex]U[/itex]. Therefore [itex]U[/itex] is open in [itex]\mathbb{R}^n[/itex].

    Why does Spivak not take this obvious route? Am I mistaken about it? Instead, he says that proof of the openness of [itex]U[/itex] needs something called the Invariance of domain theorem.
  2. jcsd
  3. Mar 19, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This step is wrong. The correct conclusion from that information is
    ... therefore U is open in U.​
  4. Mar 19, 2012 #3
    Ah, I see! And, of course, U must be open in U anyway, being a topological space; but the issue is whether U is open in [itex]\mathbb{R}^n[/itex]. Thanks, Hurkyl.
  5. Mar 21, 2012 #4
    I still don't get it. The IOD theorem is of the form "(A&B) implies C". Since f is a homeomorphism, B and C are true (f is 1-1 and continuous, and f is a homeomorphism). But this says nothing about A (U is open in R^n). (A&B) implies C" is consistent with A being true or false.

    As an aside, does the theorem assume that f is onto and hence invertible, or is this implied by the antecedents?
  6. Mar 22, 2012 #5
    Replace C with C&D. I believe you're correct that D is trivial, and it was strange to include it. However, C is still certainly nontrivial (the openness of V in ℝn).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook