Intermediate Value Theorem
Suppose \(a < b\) and that \(f: [a,b] \to \mathbb{R}\) is continuous. If \(y_0\) lies between \(f(a)\) and \(f(b)\), then there is an \(x_0 \in (a,b)\) such that \(f(x_0) = y_0\).

Proof

Suppose without the loss of generality that \(f(a) < y < f(b)\). Consider \(F(x) = y - f(x)\). Then

$$ \begin{align*} F(a) &= y - f(a) > 0 \\ F(b) &= y - f(b) < 0 \\ \end{align*} $$

We know \(F\) is continuous on \([a,b]\). Let \(I_1 = [a,b]\) and let \(x_1 = \frac{a+b}{2}\). If \(F(x_1) = 0\), then we’re done. Otherwise, set \(I_2\) as follows

$$ \begin{align*} F(x_1) > 0 \quad \longrightarrow \quad I_2 &= [x_1, b] \\ F(x_1) < 0 \quad \longrightarrow \quad I_2 &= [a, x_1] \end{align*} $$

Suppose now that \(I_k = [a_k, b_k]\) has been constructed for some \(k > 1\) such that \(F(a_k) > 0\) and \(F(b_k) < 0\). Let \(x_{k} = \frac{a_k+b_k}{2}\). If \(F(x_{k}) = 0\), then we’re done. Otherwise, set \(I_{k+1}\) as follows

$$ \begin{align*} F(x_k) > 0 \quad \longrightarrow \quad I_{k+1} &= [x_k, b_k] := [a_{k+1}, b_{k+1}] \\ F(x_k) < 0 \quad \longrightarrow \quad I_{k+1} &= [a_k, b_k] := [a_{k+1}, b_{k+1}] \end{align*} $$

We know that \(F(a_k) > 0\) and \(F(b_k) < 0\) so by construction we know that \(F(a_{k+1}) > 0\) and that \(F(b_{k+1}) < 0\). Observe now that

$$ \begin{align*} I_k \subset \ldots \subset I_3 \subset I_2 \subset I_1 \end{align*} $$

We also know that

$$ \begin{align*} |I_{k+1}| = \frac{|I_{k}|}{2} = \frac{b - a}{2^{k-1}} \end{align*} $$

which implies that

$$ \begin{align*} \frac{b - a}{2^{k-1}} = 0 \quad \text{ as } \quad k \to \infty \end{align*} $$

Thus by the Nested Interval Property, the intersection of all the closed intervals \(I_k\) is non-empty. Furthermore, the intersection contains one point so

$$ \begin{align*} \bigcap_{k = 1}^{\infty} I_k = \{x_0\} \end{align*} $$

Moreover, \(a_k \leq x_0 \leq b_k\) for all \(k\). Hence, \(a_k\) and \(b_k\) converge to \(x_0\). But recall that \(F\) is continuous. Therefore, by the Sequential Characterization of Continuity Theorem,

$$ \begin{align*} F(a_k) \rightarrow F(x_0) \quad \text{ and } \quad F(a_k) \rightarrow F(x_0) \end{align*} $$

But \(F(a_k) > 0\) for all \(k\) so \(F(x_0) \geq 0\) and \(F(a_b) < 0\) for all \(k\) so \(F(x_0) \leq 0\). Hence, \(F(x_0) = 0 = y - f(x_0)\) and \(f(x_0) = y\) as desired. \(\ \blacksquare\)

References

  • Lecture Notes by Professor Chun Kit Lai