Problem: The interval $[a,b]$ $\subset$ $\mathbb{R}$ and interval $[0,1]$ $\subset$ $\mathbb{R}$ are homeomorphic in their standard topologies.
Definition of Homeomorphism: We call the underlying bijection of a topological equivalence $h : X \to Y$ a homeomorphism. We say $X$ and $Y$ are homeomorphic.
This is how I showed it.
Let us define a function $f:[a,b]$$\to$$[0,1]$ by $f(x)$ = $\frac{x-a}{b-a}$, then we know that clearly $f(x)$ is well defined. Let, $f(x_1)$ = $f(x_2)$, then $\frac{x_1-a}{b-a}$ = $\frac{x_2-a}{b-a}$, therefore from that we get that $x_1-a$ = $x_2-a$ (when $b \neq a$). Therefore, $x_1$ = $x_2$. Thus, $f$ is injective or one-to-one. Let, $y$ $\in$ $[0,1]$ be any element and $f(x) = y$, then
$\implies \frac{x-a}{b-a}$ = $y$
$\implies x = (b-a)y + (x-a)$ = $a(1-y)+by$ $\in$ $[a,b]$. As, $[a,b]$ is a convex set, so each element in $[0,1]$ has a pre-image in $[a,b]$ as $a(1-y)+by$ is a pre-image of $y$. Now, clearly $f$ is surjective. Hence, it is bijective.
We now prove that $f$ is continuous. Let, $x_n$ be a sequence in $[a,b]$, such that $x_n \to x$. Then, we want to prove that $f(x_n) \to f(x)$. Now, $f(x_n) = \frac{x_n-a}{b-a}$ $\to$ $\frac{x-a}{b-a}$ [Since, $x_n \to a$]. Therefore, we can see that, $f(x_n) \to f(x)$. Therefore, it is continuous.
Since, $[a,b]$ is a compact set (can I say it has standard (order) topology $b/c$ its sub-basis consists of positive and negative open rays or $(-\infty,a) \cup (b,\infty)$ ?) in standard topology, and also $[0,1]$ is $T_2$ in usual topology, we know that any continuous bijection from a compact space to a topological space is homeomorphism. Hence, $f$ is homeomorphism and $[a,b]$ and $[0,1]$ are homeomorphic. (Proved)
Please note my prof hasn't done compact sets so far so it will be a problem if I use that concept while proving it in class. Hence, it would be great if someone verifies my proof, helps me on notation and also explains me to show its homeomorphic without using compact set at the end.
Appreciate your continuous support. 🙂
Best Answer
Your proof basically does the following:
Define the map $f:[a,b]\to [0,1]: x\mapsto \frac{x-a}{b-a}$.
Then note that $f$ is bijective and we have (this follows from your surjectivity proof)
$f^{-1}: [0,1]\to [a,b]: x\mapsto a(1-x)+bx$.
Then you showed that $f$ is continuous using sequences but I'm not a fan of this argument. What you essentially used is that addition and scalar multiplication are continuous, so I recommend to use these facts to conclude instantly that both $f$ and $f^{-1}$ are continuous. This also lets you avoid the compactness argument.
Also these sets are Hausdorff for the simple reason they are metrizable. No need to invent complex arguments involving the order topology.