Proof.

1.

Suppose \(f\) is not bounded above. Then, for any \(\, n\in \mathbb {N},\, \exists \, x_n \in [a,b]\,\ni \, \begin {vmatrix} f(x_n)\\ \end {vmatrix}> n\hspace {0.2cm} \forall \, n\in \mathbb {N}\hspace {0.2cm} \cdots \cdots \hspace {0.3cm} (I)\). Thus, there is a sequence \(\big \{x_n\big \}^{\infty }_{n = 1}\) in \([a,b]\) which satisfies \((I)\ni x_n \in [a,b]\hspace {0.2cm} \forall n \in \mathbb {N}\) and \(\big \{x_n\big \}^{\infty }_{n = 1}\) is bounded above. Therefore, by Bolzano-Weiestrass theorem the sequence \(\big \{x_n\big \}^{\infty }_{n = 1}\) has a convergent subsequence, say \(\big \{x_{n_k}\big \}^{\infty }_{k = 1}\).
Let \(\displaystyle {\lim _{k\rightarrow \infty } x_{n_k} = x}\). Then \(x\) is a limit point of \([a,b]\). Since \([a,b]\) is closed, \(x\in [a,b]\). Putting \(n = n_k\) in \((I)\), yields \(\begin {vmatrix} f(x_{n_k})\\ \end {vmatrix} > n_k > K\hspace {0.2cm} \forall \, K \in \mathbb {N}\, \implies \, \Big \{f(x_{n_k})\Big \}\) is not bounded.
\(\implies \, \Big \{f(x_{n_k})\Big \}\) is not convergent. This contradiction since \(f\) is continuous on \([a,b]\). Hence \(f\) is bounded on \([a,b]\).

2.

Let \(M = \sup \big \{f(x): \, x\in [a,b]\big \}\). Then in view of the completeness axiom, for each \(n\in \mathbb {N}\) an \(x_n \in [a,b]\) and hence a sequence \(\big \{x_n\big \}^{\infty }_{n = 1}\) in \([a,b]\ni M - \dfrac {1}{n} < f(x_n) < M \hspace {0.2cm} \forall \, n \in \mathbb {N}\). Thus, \(M - \dfrac {1}{n} < f(x_n) \leq M < M +\dfrac {1}{n}\hspace {0.2cm}\forall \, n \in \mathbb {N}\) and therefore \(\displaystyle {\lim _{n\rightarrow 0} f(x_n) = M}\hspace {0.3cm}\cdots \cdots \hspace {0.3cm}(II)\)
Further, note that the sequence \(\big \{x_n\big \}^{\infty }_{n = 1}\) being in \([a,b]\) is bounded. Therefore, by Bolzano-Weietrass theorem, the sequence \(\big \{x_n\big \}^{\infty }_{n = 1}\) has a convergent subsequence, say \(\big \{x_{n_k}\big \}^{\infty }_{k=1}\).
Let \(\displaystyle {\lim _{k\rightarrow \infty } x_{n_k} = \alpha }\). Then \(\alpha \in [a,b]\). Since \(f\) is continuous at \(\alpha \), \(\, \displaystyle {\lim _{k\rightarrow \infty }f( x_{n_k}) = f(\alpha )}\). On the other hand, in view of \((II)\) \(\,\displaystyle {\lim _{k\rightarrow \infty } x_{n_k} = M}\). Now, since a sequence cannot converge to two different limits, conclude that \(M = f(\alpha )\).
\(\implies \, \) The proof of \(f(\beta ) = \inf \big \{ f(x):\, x \in [a,b]\big \}\) is similar.

However, note the following:

1.

A continuous function on an open interval is not necessarily bounded. For example, the function \(f(x) = \dfrac {1}{x},\hspace {0.3cm} (0,1]\) but it is not bounded on the interval, since \(\displaystyle {\lim _{x\rightarrow 0^+} f(x) = \infty }\).

2.

A continuous function on an open interval, even if bounded does not attain its supremum and infimum. For example, the function \(f(x) = \dfrac {x}{x + 1}\hspace {0.2cm},\hspace {0.2cm} x\in (0,1)\) is continuous and bounded on \((0,1)\) but it does not attain its supremum and infimum on (0,1).

Proof. Since \(f\) is continuous at \(p\in (a,b)\), for \(\varepsilon > 0\) there exist a \(\delta > 0\, \ni \, \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \,\) whenever \(\,\begin {vmatrix} x - p\\ \end {vmatrix} < \delta \). \(\implies \, f(p) - \varepsilon < f(x) < f(p) + \varepsilon \) whenever \(x\in \big (p-\delta \, , \, p + \delta \big )\). If \(f(p) > 0\), then for \(\varepsilon < f(P)\) we have \(f(x) > 0\hspace {0.3cm} \forall x \in \big (p - \delta \, ,\, p+\delta \big )\).
Also if \(f(p) < 0\), then \(-f(p) > 0\). Therefore \(\varepsilon < - f(p)\), we have \(f(x) < 0,\hspace {0.2cm} \forall \, x \in \big (p-\delta \, , \, p + \delta \big )\).

Similar results for points of an interval are given in the next theorem.

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Proof. Suppose \(f\) is continuous at \(a\) with \(f(a)>0\); the case \(f(a)<0\) follows by applying this to \(-f\). Take \(\varepsilon = f(a)/2 > 0\) in the definition of continuity: there is \(\delta >0\) such that \[\left |f(x)-f(a)\right | < \frac {f(a)}{2} \qquad \text {whenever } \left |x-a\right |<\delta ,\ x\in [a,b].\] The left inequality gives \(f(x) > f(a) - f(a)/2 = f(a)/2 > 0\) on that neighbourhood, so \(f\) is positive there and in particular non-zero. □

Proof. Suppose that \(a<b\). We consider the set \(A = \big \{ x\in [a,b]:\, f(x) < 0\big \}\). Since \(f(a) < 0\), it follows that \(A\) is not empty. Also, since \(f(b) > 0,\,\, b\not \in A\). Then \(b\) is an upper bound of \(A\). Therefore, by completeness axiom, \(A\) has as a supremum.
Let \(P = \sup A\). Then \(P = \inf A^C\). Since \(f(a) < 0\hspace {0.2cm} \exists \, \delta > 0\, \ni f(x) < 0\hspace {0.2cm} \forall \, x\in [a, a+\delta ) \hspace {0.2cm} \implies \, [a, a+ \delta ) \subseteq A\implies a + \delta \leq P\). Since \(P = \sup A \implies a\neq P\) since \(\delta > 0\).
Also since \(f(b) >0\hspace {0.2cm} \exists \, \delta _1 >0\, \ni f(x) > 0\hspace {0.2cm} \forall x\in (b - \delta _1, b]\subseteq A^C \implies b - \delta _1 \geq P\). Since \(P = \inf A^C \implies b \neq P\). Since \(\delta _1> 0\). Further since \(\sup A = P = \inf A^C\), we can find a monotone increasing sequence \(\big \{x_n\big \}^{\infty }_{n = 1}\) in \(A\) such that \(\displaystyle {\lim _{n\rightarrow \infty } x_{n} = P}\), and a monotone decreasing sequence \(\big \{y_n\big \}^{\infty }_{n = 1}\) in \(A^C\) such that \(\displaystyle {\lim _{n\rightarrow \infty } y_{n} = P}\). But \(f\) is continuous at \(P\). Therefore we have \[\lim _{n\rightarrow \infty } f(x_n) = f(P) = \lim _{n\rightarrow \infty } f(y_n)\] \(\implies \, f(x_n) < 0\hspace {0.2cm} \forall \, n \in \mathbb {N} \) since \(x_n \in A, \hspace {0.2cm}\forall \, n \in \mathbb {N} \implies f(y_n) \geq 0\hspace {0.2cm} \forall \, n \in \mathbb {N}\), since \(y_n \in A^C \hspace {0.2cm} \forall \, n\in \mathbb {N} \implies f(P) \leq 0\) and \(f(P) \leq 0\). Hence \(f(P)=0\).

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Proof. Without loss of generality we assume that \(f(a) < f(b)\). Let \(f(a) < K < f(b)\) be given. Define, \(g(x) = f(x) - K\, ,\hspace {0.2cm} x\in [a,b]\). Then \(g\) is continuous on \([a,b]\) such that \begin {align*} g(a) & = f(a) - K < 0\\ g(b) & = f(b) - K > 0 \end {align*}

Therefore, \(\,\exists \, p\in (a,b)\, \ni \, g(p) = 0 \implies f(p) = K\). Since \(K\) was arbitrarily chosen between \(f(a)\) and \(f(b)\), the conclusion of the theorem holds.

The above theorem can be restated as follows:
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Proof. Let \(I\) be an interval and \(f\) continuous on it, and let \(u<w<v\) with \(u,v\in f(I)\). Say \(u=f(c)\) and \(v=f(d)\) with \(c,d\in I\). Since \(I\) is an interval it contains the closed interval with endpoints \(c\) and \(d\), and \(f\) is continuous there, so by the intermediate value theorem \(f\) takes the value \(w\) at some point between \(c\) and \(d\) — a point of \(I\). Hence \(w\in f(I)\).

So \(f(I)\) contains every point between any two of its points, which is exactly the definition of an interval. □

Proof. Since \(f\) is continuous on \([a,b]\). There exist points \(\alpha , \, \beta \in [a,b]\) such that \(f(\alpha ) = \inf \big \{f(x): \, x\in [a,b]\big \}\) and \(f(\beta ) = \sup \big \{f(x):\, x \in [a,b]\big \}\). Now, \(f\) is continuous on \([\alpha , \beta ] \subseteq [a,b]\). Therefore, by intermediate value theorem \(f\) assumes every value between \(f(\alpha )\) and \(f(\beta )\) i. e between the infimum and supremum of \(f\) on \([a,b]\).


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