Continuous function from set to proper subset is a contraction

Question

I want to show for part of a proof that any continuous mapping $f \colon [a,b] \to [a,b)$ must be a contraction. I'm not even sure if this is true, or how to prove it if it is.

Answer 1

Let $f:[0,1]\to [0,1)$ be given by $f(x)=\frac12\sqrt{x}$. Then the ratio $|f(x)-f(0)|/|x-0|$ grows without bound as $x\to0$.

Answer 2

If you mean that $f $ is ONTO $[0,1) $, then this is impossible, since the image of $f $ needs to be compact, and indeed $[0,1) $ is not compact.

In an elementary sense, this means as follows. Consider the sequence $\{1-\frac {1}{n}\} $. Each of its elements is in $[0,1) $ and thus has a source $\{x_n\} $. By Bolzano we get that there exists a convergent subsequent $\{x_{n_j}\}\to x $. By continuity we get [ 1=\lim 1-\frac{1}{n}=\lim f (x_{n_j})=f (\lim x_{n_j})=f (x) ] And thus 1 is in the image of $f $, which is a contradiction.

However, this works only for the specific case. Note that there is a continuous function $f$ on $[0,1]$:with image the subset $[0,\frac {1}{2}] $ given by $x\mapsto \frac{x}{2} $.