Minimum of Finitely Many Continuous Real Functions is Continuous

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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 $x \in X$:

\(\ds \map m x\) \(=\) \(\ds \min \set {\map {f_1} x, \map {f_2} x}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\paren {\map {f_1} x + \map {f_2} x} - \size {\map {f_1} x - \map {f_2} x} }\) Minimum Function in terms of Absolute Value

From Combined Sum Rule for Continuous Real Functions:

$f_1 + f_2$ and $f_1 - f_2$ are continuous.

From Absolute Value of Continuous Real Function is Continuous:

$\size {f_1 - f_2}$ is continuous.

Applying Combined Sum Rule for Continuous Real Functions 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:

\(\ds \map m x\) \(=\) \(\ds \min_i \map {f_i} x\)
\(\ds \) \(=\) \(\ds \min \set {\min_{i \mathop \le N} \map {f_i} x, \map {f_{N + 1} } x}\)

By the induction hypothesis:

$\ds \min_{i \mathop \le N} f_i$

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.

$\blacksquare$