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Topology/Analysis: Homeomorphism

Posted by jessepfrancis on Dec 16, 2020Dec 21, 2020

Definition

If $f:M_1\to M_2$ is

one-one and onto (1-1 correspondence/bijection)
continuous
$f^{-1}$ is continuous

Then we call $f$ a homeomorphism between two metric spaces $M_1,M_2$ . The metric space $M_1,M_2$ are said to be homeomorphic.

Properties and Theorems

If f is a homeomorphism between metric spaces ,
- The set G⊂ $M_1$ is open if and only if the image f(G)⊂ $M_2$ is open.
- The set F⊂ $M_1$ is closed if and only if the image f(F) is closed.

Examples

Consider metric spaces <[0,1],ρ>,<[0,2],ρ> and the function f:[0,1]→[0,2] given by f(x)=2x is a homeomorphism from [0,1] onto [0,2].
Consider f(x)=log⁡ x. Then f is a homeomorphism of (0,∞) onto $\mathbb{R}$

Counterexamples

(0,1) and [0,1] are not homeomorphic.
$\mathbb{R}$ and $\mathbb{R}_d$ are not homeomorphic

(more examples and counterexamples)

Simple Quizzes

1. Suppose $f:\mathbb{R}\to M_1$ is a homeomorphism. Which of the following is true?

f([0,1]) is open in $M_1$

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism.

f([0,1]) is closed in $M_1$

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism.

f([0,1]) is closed and open in $M_1$

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism. Similarly, if G is open in $M_1$ , then $f^{-1}(G)$ is open in $\mathbb{R}$ since f is a homeomorphism.

f([0,1]) is open but not closed in $M_1$

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism.

2. Suppose $f:\mathbb{R}\to\mathbb{R}_d$ , $f(x) = x$ , where $\mathbb{R}_d$ is real numbers with discrete metric. Then

f is continuous

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism.

$f^{-1}$ is continuous

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism.

f is a homeomorphism

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism. Similarly, if G is open in $M_1$ , then $f^{-1}(G)$ is open in $\mathbb{R}$ since f is a homeomorphism.

None of the above

If F is closed in $\mathbb{R}$ , then f(F) is closed in $M_1$ since f is a homeomorphism.

3. Which of the following is\are homeomorphisms? (Multiple choices may be correct)

$f:[0,1]\to[0,1], f(x) = x$

$f:[0,1]\to[0,\frac12], f(x) = \frac{x}2$

$f:[0,1]\to[0,2], f(x) = 2x$

$f:[0,1]\to[0,2], f(x) = x$

$f^{-1}$ is not a funciton

4. Which of the following is true? (Multiple answers may be true)

$(0,\infty)$ and $\mathbb{R}$ are homeomorphic

Check out the f(x) = log x example above.

(0,2) and [0,2] are homeomorphic

(0,1) and $\mathbb{R}$ are homeomorphic

Read Math.SE Article

None of the above

Two options are correct

Memory Trigger

Two spaces are homeomorphic if they share similar structure – maybe this image will help you remember the concept better!

Published by jessepfrancis

Christian. Artist. Mathematician. Programmer. Teacher. Visit https://mathematicos.in/aboutjesse/ for my full profile. View all posts by jessepfrancis

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Pingback: Real Analysis: Homeomorphism | Lecture Diaries

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