Supremum of Continuous Bounded Real-Valued Function on Everywhere Dense Subset

Theorem

Let $\struct {X, \tau}$ be a topological space.

Let $\DD$ be an everywhere dense subset of $\struct {X, \tau}$.

Then:

$\ds \sup_{x \mathop \in \DD} \map f x = \sup_{x \mathop \in X} \map f x$

Since $\DD \subseteq X$ we have:

$\ds \sup_{x \mathop \in \DD} \map f x \le \sup_{x \mathop \in X} \map f x$

Conversely, from the definition of supremum, for each $\epsilon > 0$ there exists $y \in X$ such that:

$\ds \sup_{x \mathop \in X} \map f x - \frac \epsilon 2 \le \map f y \le \sup_{x \mathop \in X} \map f x$

Since $f$ is continuous, there exists an open set $U$ such that:

$\ds \size {\map f y - \map f w} < \frac \epsilon 2$

hence:

$\ds \map f y - \frac \epsilon 2 < \map f w < \map f y + \frac \epsilon 2$

for $w \in U$.

Since $\DD$ is everywhere dense, we have $\DD \cap U \ne \O$.

Picking $w \in \DD \cap U$, we have:

$\ds \map f w > \map f y - \frac \epsilon 2 \ge \sup_{x \mathop \in X} \map f x - \epsilon$

while:

$\ds \map f w \le \sup_{x \mathop \in \DD} \map f x$

since $w \in \DD$.

Hence for each $\epsilon > 0$ we have:

$\ds \sup_{x \mathop \in X} \map f x \le \sup_{x \mathop \in \DD} \map f x + \epsilon$

Taking $\epsilon \to 0^+$, we then obtain:

$\ds \sup_{x \mathop \in X} \map f x \le \sup_{x \mathop \in \DD} \map f x$

Hence we have:

$\ds \sup_{x \mathop \in \DD} \map f x = \sup_{x \mathop \in X} \map f x$

$\blacksquare$