Proof. We first note that if \(x\in I\) and \(x<c\) then \(f(x) \leq f(c)\). Thus the set \(\big \{f(x):\, x\in I\, ,\, x < c\big \}\) is bounded above by \(f(c)\) and is non-empty since \(c\) is not an end point of \(I\). Hence supremum exists for the set \(\big \{f(x):\, x\in I\,,\, x < c\big \}\). Let this supremum be \(L\). If \(\varepsilon > 0\) is given then \(L- \varepsilon \) is not an upper bound of this set. Hence there exists \(y_{\varepsilon }\in I,\, y_{\varepsilon } < c\) such that \(L - \varepsilon < f(y_{\varepsilon }) \leq L\). Since \(f\) is increasing, we deduce that if \(\delta (\varepsilon ) = c- y_{\varepsilon }\) and if \(0 < c - y < \delta (\varepsilon )\) then \(y_{\varepsilon } < y < c\) so that \(L-\varepsilon < f(y_{\varepsilon }) \leq f(y) \leq L\). Therefore \(\begin {vmatrix} f(y) - L\\ \end {vmatrix} < \varepsilon \) whenever \(0 < c - y < \delta (\varepsilon )\). Since \(\varepsilon > 0\) is arbitrary, we conclude that \(\displaystyle {\lim _{x\rightarrow c^-}f = \sup \big \{f(x):\, x\in I\, , \, x<c\big \}}\).

Prove (ii) in a similar way.

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Proof. By the theorem the one-sided limits \[L^{-} = \lim _{x\rightarrow c^{-}}f(x) = \sup _{x<c}f(x),\qquad L^{+} = \lim _{x\rightarrow c^{+}}f(x) = \inf _{x>c}f(x)\] both exist, and monotonicity gives \(L^{-}\leq f(c)\leq L^{+}\). The function is continuous at \(c\) precisely when both one-sided limits equal \(f(c)\), that is when \(L^{-}=L^{+}\), and the jump at \(c\) is \(L^{+}-L^{-}\geq 0\). □

Proof. Let \(c\in I\) and suppose first that \(c\) is not the right endpoint. The set \(\left \{f(x) : x\in I,\ x>c\right \}\) is bounded below by \(f(c)\), since \(f\) is increasing, so it has an infimum; call it \(L\). For \(\varepsilon >0\) there is \(x_1>c\) in \(I\) with \(f(x_1)<L+\varepsilon \), and then for every \(x\) with \(c<x<x_1\) monotonicity gives \[L \leq f(x) \leq f(x_1) < L+\varepsilon .\] Taking \(\delta = x_1-c\) shows \(\lim _{x\rightarrow c^{+}}f(x) = L\). The left-hand limit is the supremum of \(\left \{f(x):x<c\right \}\) by the same argument reversed. □

Proof. We consider the case \(f\) is strictly increasing. Since \(f\) is continuous and \(I\) is an interval, the preservation of interval theorem imply that \(J = f(I)\) is an interval. Moreover since \(f\) is strictly increasing on \(I\), it is injective on \(I\); therefore, the function \(g:\, J\longrightarrow \mathbb {R}\) inverse to \(f\) exists. We claim that \(g\) is strictly increasing. To see this, suppose that \(y_1,\, y_2\in J\) with \(y_1 < y_2\), then \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\) for some \(x_1\, ,\, x_2\in I\). We must have \(x_1 <x_2\), otherwise \(x_1\geq x_2\) which implies that \(y_1 = f(x_1) \geq f(x_2)=y_2\). i.e \(y_1 \geq y_2\) which is a contradiction since \(y_1 < y_2\) by assumption. Therefore \(g(y_1) = x_1 < x_2 = g(y_2)\). i.e \(g(y_1) < g(y_2)\) showing that \(g\) is increasing. Proving the claim.
Since \(y_1\) and \(y_2\) are arbitrary elements of \(J\) with \(y_1 < y_2\), we conclude that \(g\) is strictly increasing on \(J\). The fact that \(g\) is continuous on \(J\) is a consequence of the fact that \(g(I) = I\) is an interval. Indeed, \(c\in J\), then the jump of \(g\) at \(c\) is nonzero so that \(\displaystyle {\lim _{y\rightarrow c^-}\, g < \lim _{y\rightarrow c^+}\, g}\).
If we chose any number \(x\neq g(c)\) with \(\displaystyle {\lim _{y\rightarrow c^-}\, g < x < \lim _{y\rightarrow c^+}\, g}\,\) then \(x\) has the property that \(x\neq g(y)\) for any \(y\in J\). Hence \(x\not \in I\) contradicting \(I\) being an interval. Therefore we conclude that \(g\) is continuous on \(J\).

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