Show that $\left [ 0,1 \right )$ is homeomorphic to $\left ( 0,1 \right ]$. 
Question:
Show that $\left [ 0,1 \right )$ is homeomorphic to $\left ( 0,1 \right ]$.

I know what the definition of homeomorphism is. However, the prove seems to require more advanced theorems.
Would

Theorem:
  Let X and Y be homeomorphic topological space and let $x \in X$. Then there exists $y \in Y$ such that $X \setminus \left \{ x \right \}$ is homeomorphic to $Y \setminus \left \{ y \right \}$ suffice?

Can someone give me a hint?
Thanks in advance.
 A: Well think of how you would translate the first to the second, you can flip it $f(x)=-x$ and then move it $g(x)=1+f(x)=1-x$
A: To answer your second question: No, that theorem is not sufficient. You can of course get $[0,1)$ by taking away the element $1$ from $[0,1]$, and of course $[0,1]$ is trivially homeomorphic to $[0,1]$. But then, all the theorem tells you is that you can take an element away from $[0,1]$ so that you get a set homeomorphic to $[0,1)$ — but you already knew that: Taking the element $1$ away does that. Nowhere does the theorem say that $y\ne x$. Nor does the theorem say that there exists more than one such element. And indeed, in the general case there may not exist a second one. Consider taking away $0$ from $[0,1)$ to get $(0,1)$; there's no other element you can take away from $[0,1)$ to get a set homeomorphic to $(0,1)$.
The easiest (and I believe, for this specific problem the only) way to prove the sets to be homeomorphic is to explicitly construct a homeomorphism.
A: Just try drawing their cartesian product [0,1)x(0,1] in the plane and connecting the upper left corner with the lower right corner. That subset of their product is a continuous bijection with a continuous inverse. 
Always reason about these problems geometrically. You can't pull proofs out of thin air. They have to come from somewhere. Use your intuition about the concepts. 
