Why does taking the supremum preserve inequality in the uniform metric?

[Math Amateur]

TL;DR

This thread involves a proof that the uniform metric is indeed a metric ...

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help with an aspect of the proof of Theorem 11.1.11 ...

Garling's statement and proof of Theorem 11.1.11 reads as follows:

Near the end of Garling's proof above we read the following:

" ... ... Suppose that $f,g,h \in B_X(S)$ and that $s \in S$. Then

$d(f(s), h(s)) \le d(f(s), g(s)) + d(g(s), h(s)) \le d_\infty (f, g) + d_\infty (g, h)$ ... ... ... (1)

Taking the supremum, $d_\infty (f, h) \le d_\infty (f, g) + d_\infty (g, h)$

... ... ... "Now (1) is true for arbitrary s and so it is true for all $s$ including the point for which $d(f(s), h(s))$ is a maximum ... if a maximum exists ...

But in the case where a maximum does not exist ... how do we know that taking the supremum preserves inequality (1) ...Hope someone can help ...

Peter

 

[pasmith]

Line (1) asserts that $d_\infty(f,g) + d_\infty(g,h)$ is an upper bound for $\{d(f(s),g(s)) : s \in S\}$. By definition of "least upper bound" it follows that $\sup\{d(f(s),g(s)) : s \in S\} \leq d_\infty(f,g) + d_\infty(g,h)$.

 

[Math Amateur]

Thanks for the help, pasmith ...

Peter