Purchase Solution

Bijection Proof

Not what you're looking for?

Ask Custom Question

Let f : M -> N be a continuous bijection. M is compact.
Show that f is a homeomorphism.

Isn't a homeomorphism by definition a bijection? And since M is compact, will it not be true that N will be compact too?

Purchase this Solution
Solution Summary

This is a proof regarding bijection and homemorphism.

Solution Preview

yes, homeomorphisms are bijections by definition alone;
and since M is compact, f(M) = N is compact as well

to prove the claim, you also want N to be Hausdorff

in that ...

Purchase this Solution
Free BrainMass Quizzes
Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

Exponential Expressions

In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.

Probability Quiz

Some questions on probability

Multiplying Complex Numbers

This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

View More Free Quizzes
Related BrainMass Solutions

  • Prove that there is no bijection between any set A and its power set P(A) of A.

    24567 Prove that there is no bijection between any set A and its power set P(A) of A. There is no bijection between any set A and its power set P(A) of A. For finite sets, proof is trivial since |A| = n and |P(A)| = 2^n.

  • Euler Function

    Prove that establishes a bijection between and . (In other words, if f(k)=(a,b), then gcd(k,mn)=1 if and only if gcd(a,m)=1 and gcd(b,n)=1) c. Use the results of the preceeding two exercises to complete the proof of the conjecture.

  • Cosets

    Therefore, $ is a well-defined bijection. This is a proof regarding cosets and bijections.

  • Topology : Past Exam Paper

    Proof: is open means that for any open set in , is an open set in . Now suppose is a continuous open bijection, then for any , . Then for any neighborhood , we can find an neighborhood , such that . Now we consider .

  • Equivalence Relation on the set R of all Real Numbers

    ) to the set {x: x is a real number and 0 <= x < 1}, and a step-by-step proof that that mapping is a well-defined bijection.

  • Denumerability and induction

    Show by induction that 6 divides n^3 - n for all n in N Proof: If n=1, then n^3-n=0. So 6 divides 0=n^3-n. Now suppose 6 divides n^3-n.

  • Functions, Enumeration Schemes and Bijection

    Proof. We need to prove two things, one is to prove f is one-to-one; the other is to prove f is onto. (1) f is one to one.

  • Functions : Onto and One-to-one, Bijections and Function Composition 'f o g'

    Proof: f:A->B, g:B->C are bijections. Then f^(-1): B->A, g^(-1): C->B are bijections. Then (gof)^(-1): C->A is a bijection. For any x in C, we have (gof)^(-1)(x)=y for some y in A. Then (gof)(y)=x. Then g(f(y))=x.

  • Rings with Unity, Isomorphism, Bijectiveness and Invertibility

    Proof: First, I show that is a ring. For any , we know and . Thus . So is the additive identity of . Since and , then we have . So is the multiplicative identity. It is clear that is an abelian additive group. and , then .

View More Related Solutions