- #1

kakarukeys

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if one of the subspaces is open in [tex]R^n[/tex], is the other one open in [tex]R^n[/tex] too?

I know it is trivial, but I can't see any solution.

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- #1

kakarukeys

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if one of the subspaces is open in [tex]R^n[/tex], is the other one open in [tex]R^n[/tex] too?

I know it is trivial, but I can't see any solution.

- #2

mathwonk

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for example just try to prove there is no homeomorphism from R^3 to R^4,

or even S^2 to S^3.

or even S^1 to S^2.

just try to prove that every loop on S^2 can be shrunk to a point.

- #3

Palindrom

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I don't even think it's true. But of course I've only taken one introductory class so...

- #4

mathwonk

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now ask yourself what about R^1? is an oen interval homeomorphic to something else in the real line?

now think about R^2. let D be an open disc centered at 0. a homeo maps 0 say to 0. now what does it do to a small circle around 0? does it map to a small loop arounmd the image 0? if not what else could happen?

if it mapped to a loop that did not surround 0, then as you shrink the circle wouldn't it shrink along a cone of circles down to 0 and become non injective at some point?

i.e. i claim an injective continuous map from an open disc in R^2, to R^2 must map to an open nbhd of every image point.

now apply the same reasoning to an open ball in R^3 mapping to R^3. etc...

- #5

kakarukeys

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I understand the argument.

I don't see how it helps to solve the problem.

I don't see how it helps to solve the problem.

- #6

mathwonk

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- #7

kakarukeys

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yes, how do you arrive at this statement from the previous one (about the loops around 0)

- #8

selfAdjoint

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kakarukeys said:

if one of the subspaces is open in [tex]R^n[/tex], is the other one open in [tex]R^n[/tex] too?

I know it is trivial, but I can't see any solution.

Since it's a homeomorphism onto, therefore continuous onto, take the image subspace and suppose it's open, then by continuity its pre-image under the function is open too. But that's the source subspace! So that proves the statement one way. What property of a homeomorphism can you use to prove it the other way?

- #9

mathwonk

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as to the OP, if you have an injective continuous map from an open disc D centered at A, to R^2, and A goes to B, then in order to show the image of the disc contains an open nbhd of B, the first observation is that it suffices to show that some circle centered at A has image which "winds around B".

once you realize this, you ask yourself what would happen if the image of every circle failed to do so, or in fact even if the image of one circle failed to do so. then the image of A and this circle, would look like B and a loop that does not have B in its interior.

now look at smaller and smaller circles around A and ask where they go? they must go to smaller and smaller circles NOT encircling B. but it is plausible to imagine that as these image circle shrink down to B without ever enclosing it, that their images must overlap one another, hence the map is not injective.

this is just an intuitive argument, and indeed this is not at all easy to prove. indeed even if we assume more, that the map is injective and continuously differentiable with invertible derivative everywhere, this is the main step in the proof of the inverse function theorem.

there however we can use the derivative to show that every point nearer to B than to the image f(C) of the boundary circle C of D, is in the image of f. i.e. take a point Q close to B. then someif f(y) is not equal to Q, then by the basic principle of derivatives vanishing at minima, the derivative of f at y is zero, a contradiction.

a nice source for the differentiable case is spivaks calculus on manifolds, inverse function theorem.

for the continuous case you coulds use the jordan curve theorem to tell you that the image of the boundary circle C is a simple closed curve f(C) which separates the plane into two connected components, both open, an inside and an outseid, and that f(C) winds exactly once around each inside point. then it follows from the principle mentioned above that f(D) must map homeomorphically onto the inside component of f(C) which is open.

- #10

sparkster

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Does this work?

Let [tex]f: A \rightarrow B[/tex],a homeomorphism.

Let [tex]g:A \rightarrow R^n[/tex] be the extension map of [tex]f[/tex].

Let [tex]h:B \rightarrow R^n[/tex] be the extension map of [tex]f^{-1}[/tex].

As extensions of codomains of contiuous functions, g and h are continuous.

Assume A is open in [tex]R^n[/tex]. Then B is open since h is continuous. Assume B is open, then A is open since g is continuous.

Let [tex]f: A \rightarrow B[/tex],a homeomorphism.

Let [tex]g:A \rightarrow R^n[/tex] be the extension map of [tex]f[/tex].

Let [tex]h:B \rightarrow R^n[/tex] be the extension map of [tex]f^{-1}[/tex].

As extensions of codomains of contiuous functions, g and h are continuous.

Assume A is open in [tex]R^n[/tex]. Then B is open since h is continuous. Assume B is open, then A is open since g is continuous.

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- #11

kakarukeys

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in order to show the image of the disc contains an open nbhd of B, the first observation is that it suffices to show that some circle centered at A has image which "winds around B".

Why??

- #12

mathwonk

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- #13

kakarukeys

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You didn't get my question.

- #14

mathwonk

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i answered the contrapositive of your question. i.e. you asked me to explain why a map sends a disc is onto a nbhd of B, if the boundary of the disc winds around B. I showed that if the maps does not hit a point, then the boundary does not wind around that point.

now suppose the boundary does wind aropund B, then it also winds around all points near B, and thus by the contrapositive of what I proved, it also maps onto all points near B.

i.e. in logic, "C implies D" is equivalent to "not D implies not C", also called "proof by contradiction".

now suppose the boundary does wind aropund B, then it also winds around all points near B, and thus by the contrapositive of what I proved, it also maps onto all points near B.

i.e. in logic, "C implies D" is equivalent to "not D implies not C", also called "proof by contradiction".

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- #15

kakarukeys

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But I have a question

Suppose I have a circle surrounding A, mapped to a loop "winding around" B.

(A mapped to B).

I agree that if the loop winds around B, then the loop winds around the nearby points of B (i.e. a small open nbhd of B).

But are these nearby points of B in the image of the map?

Remember the map is from a subspace to a subspace not the whole of R^n.

I argue that the interior of the circle might get mapped into the interior of the loop (BUT not onto the interior of the loop), so the image migh not be open.

- #16

mathwonk

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i have already argued this but maybe too briefly. here it is again:

lets prove by contradiction that the map does map onto every point the loop winds around. we prove it by contradiction. i.e. we will prove that any point not in the image of the map cannot be wound around by the loop.

suppose Q is a point that is not hit by the image of the disc. then the center misses Q, and thus any small circle near the center fails to wind around Q. now winding number about Q varies continuously as we move out along larger and larger circles (since Q is not in the image of any of the circles), but it is also an integer, so it remains constant namely zero ("winding number is a deformation invariant").

thus the winding number of the outer circle, i.e. the image loop, is the same wrt Q as the winding number of the inner "circle", i.e. the center, namely zero.

so we have argued that if Q is not in the image of the disc, then the loop whicxh is the image of the outer circle, does not wind around Q. hence vice versa, if the outer loop does wind around Q, then Q must be in the image of the disc.

these arguments are not trivial, and completely rigorous ones take a lot of work, as i said in the very beginning. but this is an intuitive argument that explains the main features.

lets prove by contradiction that the map does map onto every point the loop winds around. we prove it by contradiction. i.e. we will prove that any point not in the image of the map cannot be wound around by the loop.

suppose Q is a point that is not hit by the image of the disc. then the center misses Q, and thus any small circle near the center fails to wind around Q. now winding number about Q varies continuously as we move out along larger and larger circles (since Q is not in the image of any of the circles), but it is also an integer, so it remains constant namely zero ("winding number is a deformation invariant").

thus the winding number of the outer circle, i.e. the image loop, is the same wrt Q as the winding number of the inner "circle", i.e. the center, namely zero.

so we have argued that if Q is not in the image of the disc, then the loop whicxh is the image of the outer circle, does not wind around Q. hence vice versa, if the outer loop does wind around Q, then Q must be in the image of the disc.

these arguments are not trivial, and completely rigorous ones take a lot of work, as i said in the very beginning. but this is an intuitive argument that explains the main features.

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- #17

kakarukeys

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Clear

Another question, your proof seems to depend on circles and loops.

take the homeomorphism to be from an open disc of [tex]R^n[/tex] to a subspace of [tex]R^n[/tex]

How do you prove that the other subspace is open? As you can see, that the boundary of the

open disc is not in the domain of the homeomorphism, we can't say it is mapped to a loop in the codomain.

Although you can prove smaller open discs which boundaries are contained inside the domain are mapped to open sets of [tex]R^n[/tex], does that prove what we want?

- #18

mathwonk

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this time you need a notion of "wrapping number" or enclosing number for an n-1 sphere about a point. but again it varies continuously as long as no one of the family of spheres hits the target point. so again we get that the smaller open disc maps onto the interior of the image of the image of the boundary.

to make this rigorous is a lot of work again.

- #19

sparkster

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ETA: Upon rereading this, it sounded kind of sarcastic. I didnt' mean for it to...I would really like to know my mistake.

- #20

mathwonk

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- #21

Timbuqtu

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For any open set [tex]U \subset R^n[/tex] and any continuous injective mapping [tex]f : U \rightarrow R^n[/tex]: the image [tex]f(U) \subset R^n[/tex] is open and [tex]f : U \rightarrow f(U)[/tex] is a homeomorphism.

- #22

mathwonk

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you are right of course. do you know brouwer's original method of proof?

- #23

Timbuqtu

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From this it can be shown that: if [tex]X,Y \subset R^n[/tex] closed and [tex]f : X \rightarrow Y[/tex] a homeomorphism, then [tex]f(\partial X)=\partial Y[/tex] (where [tex]\partial[/tex] denotes boundary).

Then we might apply this to a small closed ball [tex]B[/tex] around a point [tex]x[/tex] in the domain [tex]U[/tex] with [tex]x \in B - \partial B[/tex]. If we show that [tex]f:B \rightarrow f(B)[/tex] is a homeomorphism, we see that x maps to the interior of [tex]f(B) \subset f(U)[/tex], showing that f(U) is open.

- #24

mathwonk

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but in general i believe brouwers fix point theorem is rather easy in comparison to invariance of domain. (which of course only means i understand one better than the other.)

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