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I am trying to understand a proof, I don't understand one of the steps. could someone shed some light why the following is true.

if $f(x)$ is a non-increasing and continuous function and $a>b$ then $\{x\colon f(x) > f(b) \} \subset \{x\colon f(x) > f(a) \} $

i don't see why this is true. any insights ?

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    I have edited the Q from f(x)>a to f(x)>f(a).... BTW it is common to say "$f$ is decreasing" to mean (x>y\implies f(x)\leq f(y)\;), following the terminology of the Bourbaki books. And to say "$ f$ is strictly decreasing" to mean (x>y\implies f(x).... Many people do not like "$f$ is non-increasing" as it might be interpreted as "$f$ is not an increasing function", which applies to any non-monotonic function.2017-09-08

1 Answers 1

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To say that $f$ is nonincreasing means that for all $x,y$ in the domain, if $x \leq y$ then $f(x) \geq f(y)$.

You are given that $b < a$, so $f(b) \geq f(a)$.

Finally if $A \leq B$ are two real numbers, then for any function $f$, the set of $x$ in the domain such that $f(x) > B$ is a subset of the set of $x$ in the domain such that $f(x) > A$, just because any number which is greater than $B$ is also greater than $A$.

Note that continuity of $f$ was not used here.