If I have a continuous function $f$, is it true in general that:
$$ \sup_{x \in S} f(x) < \sup_{x \in X} f(x) $$
for a subset $S \subset X$?
Is it true in general for a continuous function $f$ and a non-empty set $X$, $\sup_{x \in S} f(x) < \sup_{x \in X} f(x)$ for a subset $S \subset X$?
0
$\begingroup$
real-analysis
elementary-set-theory
-
2No. What if $f$ is constant. It needs to be $\leq$. – 2017-02-09
1 Answers
2
No; but, if you change it to the following, then yes: $$ \sup_{x\in S}f(x)\leq \sup_{x\in X}f(x). $$
To prove it: note that you need only show that $\sup_{x\in X}f(x)$ is an upper bound on $\{f(x)\mid x\in S\}$; and, this is trivially the case, because $\sup_{x\in X}f(x)$ is by definition an upper bound on $\{f(x)\mid x\in X\}\supseteq \{f(x)\mid x\in S\}$.