If $f$ and $g$ are both continuous, then the functions $\max\{f,g\}$ and $\min\{f,g\}$ are also continuous

Question

Let $X$ be a subset of $\textbf{R}$, and let $f:X\to\textbf{R}$ and $g:X\to\textbf{R}$ be functions. Let $x_{0}\in X$. Then if $f$ and $g$ are both continuous at $x_{0}$, then the functions $\max\{f,g\}$ and $\min\{f,g\}$ are also continuous at $x_{0}$.

MY ATTEMPT

Once one knows the identities \begin{align*} \max\{a,b\} = \frac{a + b + |a-b|}{2}\quad\wedge\quad\min\{a,b\} = \frac{a + b - |a-b|}{2} \end{align*}

we can argue the continuity of the functions $\max\{f,g\}$ and $\min\{f,g\}$ as a consequence of composition of continuous functions.

My question is: is there another way to prove it without appealing to such identities? (Or one way which deduces them at least).

EDIT

As commented by @GEdgar, the continuity of $|\cdot|$ has already been proved.

Answer 1

One way of deduce these identities:

It is easy to see that $\max\{f,g\} + \min\{f,g\} = f+g$ and that $\max\{f,g\} - \min\{f,g\} = |f-g|$. Adding both equalities and dividing by $2$ we obtain the first identity, and substracting the second from the first one, and again dividing by $2$, we obtain the second identity.

Answer 2

If $f(x_0) > g(x_0)$ then let $\epsilon = \frac {f(x_0) - g(x_0)}2$. Then as $f,g$ are continuous there are $\delta_1$ and $\delta_2$ where $|x- x_0| < \min(\delta_1, \delta_2) \implies |f(x) - f(x_0)| <\epsilon, |g(x)-g(x_0)| < \epsilon$ which means $f(x) > f(x_0) - \epsilon = \frac {f(x_0) + g(x_0)}2$ and $g(x) < g(x_0) + \epsilon = \frac {f(x_0) + g(x_0)}2 < f(x)$.

So $\min{f,g}(x) = g(x)$ which is continuous at $x$ and $\max{f,g}(x) =f(x)$ which is continuous at $x$.

Equivalent argument thing if $f(x_0) < g(x_0)$.

And if $f(x_0) = g(x_0)$ then for any $\epsilon >0$ there are $\delta_1, \delta_2$ so that $|x-x_0| < \min(\delta_1, \delta_2)$ then $|f(x)- f(x_0)| = |f(x) - \min{f,g}(x_0)| < \epsilon$ and $|g(x)- g(x_0)| = |g(x) - \min{f,g}(x_0)| < \epsilon$ so $|\min{f,g}(x) - \min{f,g}(x_0)| < \epsilon$ and $|\max{f,g}(x) - \max{f,g}(x_0)| < \epsilon$. So $\min{f,g}$ and $\max{f,g}$ are continuous.

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All said and done my idea was the first thing I would have thought of...

But your idea is clever and easier. Aziff00 is perhaps the easiest and cleverest but it is essentially yours.