Question: I am solving the following problem:

> 
> Let $f(x) =\sqrt{x}$. Find a formula for a function $g$ whose graph is obtained from $f$ from the given sequence of transformations: 
> 
> 
> * shift right $3$ units
> * horizontal shrink by a factor of $2$
> * shift up $1$ unit
> 
> 
> 

I think that$ g(x) = f(2(x-3)) + 1 = \sqrt{2x-6} + 1$, but in the answers it says $\sqrt{2x-3} + 1$, so i assume $g(x) = f(2x-3) +1$, but wouldn't that mean that the horizontal shrink was done first and afterwards the right horizontal shift?

Answer: 
Let $f\_s:\mathbb{R} \to \mathbb{R}$ be a scaling function $x \mapsto sx$ for some positive scaling factor $s$; then $f\_s$ maps the unit interval to an interval of length $s$. Then let $g\_t:\mathbb{R} \to \mathbb{R}$ be a shifting function $x \mapsto x+t$ for some shift $t$. Then to show that an arbitrary interval from $a$ to $b$, $a \neq b$, is uncountable, note that $g\_a \circ f\_{b-a}$ maps the unit interval to the interval from $a$ to $b$. $f\_s$ and $g\_t$ are both bijective, so the interval from $a$ to $b$ must have the same cardinality as the unit interval. Therefore the interval from $a$ to $b$ is uncountable. Since $a$ and $b$ are arbitrary, every interval of $\mathbb{R}$ is uncountable. Q.E.D.

Note: I'm assuming a priori that we know $\mathbb I$ is uncountable, but since you used that in your prior proof, that seems OK.
**HINT:** Find an injection from $\Bbb R$ into the interval $(0,1)$. Then show there is a bijection between any two intervals $(a,b)$ and $(c,d)$.