Do you know Topology?

Probability Level 4

How many statements below are true?

1.- There exists a homeomorphism (bijective continuous function) from [-1,1] to (-1,1)

2.- There exists a homeomorphism from $\mathbb {R}^2$ to $\mathbb {R}$

3.- There exists a homeomorphism from (-1,1) to $\mathbb {R}$

4.- There exists a homeomorphism from $\mathbb {R}^2$ to $\mathbb {R}^2$ - {0}

Details and assumptions

The toplogy in each space,subspace is the usual or euclidean topology

0 4 3 2 1

1 solution

Mark Hennings
Dec 10, 2015

We need the key fact that if $f:X \to Y$ is a homeomorphism, then so is its restriction $f_0 : X \backslash\{x\} \to Y \backslash \{f(x)\}$ for any $x \in X$ . Connectedness and simple connectedness are topological properties (invariant under homeomorphism).

If $-1$ is removed from $[-1,1]$ , the resulting space $(-1,1]$ is still connected. Removing any point from $(-1,1)$ disconnects it. Thus no homeomorphism exists for #1.

Removing $(0,0)$ from $\mathbb{R}^2$ leaves the set connected. Removing any point from $\mathbb{R}$ disconnects it. Thus no homemorphism exists for #2.

The function $f(x) = \tanh x$ is a homeomorphism from $\mathbb{R}$ to $(-1,1)$ .

While $\mathbb{R}^2$ is simply connected, $\mathbb{R}^2 \backslash \{0\}$ is not. No homeomorphism exists in #4.

Nice... For 1, we can also say that [-1,1] is a compact set and (-1,1) is not a compact set, and for 4, we can also say that their fundamental groups are not the same. Thank you for your solution.

Guillermo Templado - 5 years, 6 months ago

Of course, the difference between the two fundamental groups in #4 is an immediate consequence of the fact that one space is simply connected, while the other is not!

Mark Hennings - 5 years, 6 months ago

Do you know Topology?

1 solution

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