Let $f$ be a strictly increasing function on an open interval. Is it enough to prove continuity by proving that $f$ is left continuous?

I'm inclined to think so, but haven't come up with a reason yet why this cannot be the case.

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    $\begingroup$ $$f(x) = \begin{cases}\quad x &, x \in (0,1]\\ x+1 &, x \in (1,2). \end{cases}$$ $\endgroup$ – Daniel Fischer Mar 9 '14 at 21:47

No; as Daniel Fischer's example shows, left continuity and monotonicity do not imply continuity.

A necessary and sufficient condition for a strictly increasing function to be continuous on an open interval $I$ is that $f$ satisfies the intermediate value theorem on $I$, i.e. if $a < b$ and $f(a) < r < f(b)$, then $f(c) = r$ for some $c \in (a,b)$.


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