prove that $[a,b)\cong [c,d)\cong (a,b]\cong(c,d] $ with a<b and c<d and a,b,c,d any real numbers

Question

$$[a,b)\cong [c,d)\cong (a,b]\cong(c,d] $$

this is my guess: So first I'm trying to come up with an interval and a function that will be homeomorphic for each of the intervals and then I'll use the transitivity property to show that they're all homeomorphic to each other.

But i cannot think of a function that works, i have tried on the interval $[0,1)$ and $[0,1]$

and here are some functions that find interesting and close enough to give an idea

$$ h(x)= \frac{x+c}{1-x} $$ as a failed atempt to prove an homeomorphism between $[0,1)$ $\mapsto$ $[c,d)$

any help to find the interval and the function would be awesome

tks in advance

Answer 1

Just stretch and slide, stretch and slide.

Match the roundy bits to the to the roundy bits and the square bits to the square bits. So $[a,b) \to [c,d)$ will be $x \mapsto \frac {x - a}{b-a}\times (d-c) + c$...

And $[c,d) \to (a,b]$ will be $x \mapsto \frac {x-c}{d-c}\times (a-b) + b$

And $(a,b] \to (c,d]$ will be $x \mapsto \frac {x-b}{a-b}\times (c-d) + d$

And $(c,d] \to [a,b)$ will be $x \mapsto \frac {x-d}{c-d}\times (b-a) + a$

Answer 2

For $$ f: [0,1) \to [c,d) $$ try $$ f(x) = (d-c)x+c $$

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

You can find other cases easily.