Minimum of Finitely Many Continuous Real Functions is Continuous

Theorem
Let $n \ge 2$ be a natural number.

Let $X \subseteq \R$.

Let $f_1, f_2, \ldots, f_n$ be functions $X \to \R$.

Define the function $m : X \to \R$ by:
 * $\ds \map m x = \min_i \map {f_i} x$

for all $x \in X$.

Then $m$ is continuous.

Proof
We proceed by induction.

For all natural numbers $n \ge 2$, let $\map P n$ be the proposition:
 * for every collection of $n$ functions $f_1, f_2, \ldots, f_n : X \to \R$, $m$ is continuous.

Basis for the Induction
Take $n = 2$.

We have, for each $x \in X$:

From Combination Theorem for Continuous Functions: Combined Sum Rule:
 * $f_1 + f_2$ and $f_1 - f_2$ are continuous.

From Absolute Value of Continuous Real Function is Continuous:
 * $\size {f_1 - f_2}$ is continuous.

Applying Combination Theorem for Continuous Functions: Combined Sum Rule again, we obtain:
 * $m$ is continuous.

So $\map P 2$ is true.

This is our base case.

Induction Hypothesis
Suppose that:
 * for every collection of $N$ functions $f_1, f_2, \ldots, f_N : X \to \R$, $m$ is continuous.

That is:
 * $\map P N$ is true for some $N$.

We aim to show that:
 * $\map P {N + 1}$ is true.

Induction Step
For each $x \in X$, we have:

By the induction hypothesis:
 * $\ds \min_{i \le N} f_i$

is continuous.

Since $m$ is the minimum of two continuous functions, it is continuous by the base case.

So $\map P N \implies \map P {N + 1}$ and the result follows by the Principle of Mathematical Induction.