If $w$, $x$, $y$, and $z$ are real numbers with $w < x$ and $y < z$, is the cardinality of the closed interval $[w,x]$ the same as that of $[y,z]$?

Question

My reasoning is yes. I tried to draw a few example functions and based on my workings, think that the answer should be yes but I couldn't figure out how exactly I should mathematically prove the fact. Any hints?

Answer 1

Hint: Consider the case where $[w,x] = [0,1]$ and $f\colon [0,1] \to [y,z]$ is defined by $$f(t) = tz + (1-t)y$$

Answer 2

Yes. The most basic method to prove two sets having the same cardinality is by constructing a bijection.

For that, define a function f mapping [w,x] to [y,z]. A sufficient condition for f to be a bijection is simply that f be continuous and increasing.

Answer 3

HINT

You can build a linear mapping $f:[w,x]\to[y,z]$ such that $f(w) = y$ and $f(x) = z$.

Precisely, if we let $f(t) = at + b$, it results into the following system: \begin{align*} \begin{cases} aw + b = y\\\\ ax + b = z \end{cases} \end{align*}

Can you take it from here?