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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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    Should the second set be $\{x:f(x)\gt f(a)\}$? That is what Pete L. Clark's answer reasonably assumes.2010-12-20
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    Yes, I assume so. :)2010-12-20
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    no it is $\{ x\colon f(x) > a\} $ that why i am a bit confused2010-12-20
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    @beginner: It should be $\{x:f(x)>f(a)\}$2010-12-20
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    @beginner: If that is the case, there is a typo in the proof. For a simple counterexample with $a$ there instead of $f(a)$, consider $f(x)=-x$ for all $x$, $b=0$, and $a=1$.2010-12-20
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    hmmm maybe there is a mistake in my notes, but it certainly says $\{ x\colon f(x) > a\}$2010-12-20
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    Yes, it certainly is a mistake, and you now have an answer to why the corrected version is true.2010-12-20
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    thanks makes sense now, the proof actually stated $\{ x\colon f(x) > a\} \supsetneq \{x\colon f(x) > f(b) \} $ i am guessing is the same as what i originally posted2010-12-20
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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 $(x>y\implies f(x)$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

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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.