I need to show that every interval $[a,b]$ is homeomorph to $[0,1]\subset \mathbb{R}$. I've found this answer but it only deals with open sets, and I need an answer that deals with closed sets.
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The question you ask is about $[0,1]$ to $[a,b]$. The answer you link to goes from $(0,1)$ to $\mathbb R$, and your phrasing sounds like you want something from $[0,1]$ to $\mathbb R$ (which is not possible). – May 21 '16 at 17:49
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1@TokenToucan I think that the first sentence is clear ("homeomorph" should be homeomorphic, as far as I know). – Siminore May 21 '16 at 17:52
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Why don't you linearly do it: interval length should be expaded/shrunk, and $a$ must map to $0$, so $$f:[a,b] \longrightarrow [0,1]$$ $$\ \ \ \ \ \ \ \ \ x\longrightarrow \frac{x-a}{b-a}$$ ?
It has a continuous inverse.
Behnam Esmayli
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Consider $A=\{(x,0) \mid 0 \leq x \leq 1\}$ and $B = \{ (x,1) \mid a \leq x \leq b\}$. Then draw the line thru $(1,0)$ and $(b,1)$. This line intersects the vertical axis somewhere. Use straight lines from this intersection point to construct a homeomorphism of $[0,1]$ onto $[a,b]$.
Siminore
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