Show that a bijective function from $\mathbb{R}$ to $[0,\infty)$ has infinite number of discontinuities.

Question

I was trying to prove the above result. Any idea with how to approach the question will be of great help. Thank you in advance.

Answer 1

Assume that $f : \mathbb{R} \to [0, \infty)$ has only finitely many point of discontinuities, say $x_1 < \cdots < x_n$. Then by writing

$$ I_0 = (-\infty, x_1), \quad I_1 = (x_1, x_2), \quad \cdots, \quad I_{n-1} = (x_{n-1}, x_n), \quad I_n = (x_n, \infty), $$

$f$ is continuous on each $I_i$ and thus $f(I_i)$'s are disjoint open intervals. Now notice that

$$[0, \infty) \setminus \bigcup_{i=0}^{n} f(I_i)$$

is a disjoint union of $n+1$ (possibly degenerate) closed intervals, hence we must have

$$ n = \left| f(\{x_1, \cdots, x_n\}) \right| = \left| [0, \infty) \setminus \bigcup_{i=0}^{n} f(I_i) \right| \geq n+1, $$

a contradiction. (Here, $|A|$ denotes the cardinality of the set $A$.) Therefore $f$ must have infinitely many discontinuities.