text: Q:
  
  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
  
  $$[a,b)\cong [c,d)\cong (a,b]\cong(c,d] $$
  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
  
  A:
  
  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 $x \mapsto \frac {x-d}{c-d}\times (b-a) + a$
  
  