Surjective Monotone Function is Continuous
Theorem
Let $X$ be an open set of $\R$.
Let $Y$ be a real interval.
Let $f: X \to Y$ be a surjective monotone real function.
Then $f$ is continuous on $X$.
Proof
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Without loss of generality, let $f$ be increasing.
Let $c \in X$.
From Limit of Monotone Real Function: Corollary, the one sided limits of monotone functions exist:
\(\ds L^-_c\) | \(=\) | \(\, \ds \lim_{x \mathop \to c^-} \map f x \, \) | \(\, \ds = \, \) | \(\ds \sup_{x \mathop < c} \map f x\) | ||||||||||
\(\ds L^+_c\) | \(=\) | \(\, \ds \lim_{x \mathop \to c^+} \map f x \, \) | \(\, \ds = \, \) | \(\ds \inf_{x \mathop > c} \map f x\) |
and satisfy:
- $L^-_c, L^+_c \in Y$
- $L^-_c \le \map f c \le L^+_c$
Suppose that $\ds L = \lim_{x \mathop \to c} \map f x$ exists.
From Limit iff Limits from Left and Right:
- $L = L^-_c$
This leads to:
- $L \le \map f c$
Similarly:
- $L = L^+_c$
which leads to:
- $L \ge \map f c$
Hence:
- $\ds \lim_{x \mathop \to c} \map f x = \map f c$
proving continuity at $c$.
By assumption, $f$ is increasing.
Suppose $\ds \lim_{x \mathop \to c} \map f x$ does not exist.
Then from Discontinuity of Monotonic Function is Jump Discontinuity, there is a jump discontinuity at $c$.
Aiming for a contradiction, suppose $f$ has a jump discontinuity at $c$.
From Real Numbers are Densely Ordered:
- $L^-_c < y < L^+_c$
for some $y \in Y$.
By surjectivity, $y = \map f a$ for some $a \in X$.
Hence:
- $L^-_c < \map f a < L^+_c$
If $a < c$ then $\map f a \le L^-_c$.
This contradicts the previous inequality.
There is a similar contradiction if $a \ge c$.
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