Topology

Proof that (a,b) is not homeomorphic to [a,b]. Assume that they are. Thus we have a continuous, continuously invertible function f from [a,b] to (a,b). Now because f is invertible it is bijective thus f(a)=f(b) is false. Let f(a)<f(b) so that a<f(a)<f(b)<b. Not for f to be invertible we must have a c in (a,b) such that f(c) is in (f(b),b) so that f(b)<f(c). Now by the Intermediate Value Theorem there is a d in (a,c) such that f(d)=f(b). Thus f fails to be invertible if f(a)<f(b).

Easy Math Editor

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Comments

The same result holds if f(a)>f(b).

Samuel Queen - 8 years ago

I like this proof. Usually an invariant like compactness is used, but this is more Analysis based.

Philip Doi - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$