Question:
Show that all the intervals in $\mathbb{R}$ are uncountable.
I have already proven that $\mathbb{R}$ is uncountable by using the following:
Suppose $\mathbb{R}$ is countable. Then every infinite subset of $\mathbb{R}$ is also countable. This contradicts the fact that $\mathbb{I} \subset \mathbb{R}$ is uncountable. Consequently, $\mathbb{R}$ must be uncountable.
However, how can I show that ALL the intervals in $\mathbb{R}$ are uncountable?
Best 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.