How to show that if x minimizes f over S and x belongs to R, which is a subset of S, then x also minimizes f on R Please help me with this proof. Thank you.
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1If George is the shortest adult in the United States, and George lives in Los Angeles, then George is the shortest adult in Los Angeles. – André Nicolas May 10 '12 at 13:09
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@Andre: ...and Los Angeles is in United States – SBF May 10 '12 at 13:14
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2@Ilya: As opposed to Los Angeles, Mexico. – joriki May 10 '12 at 13:16
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@Jackson I am intrigued as to what exactly do you find troubling in this question. Are you having trouble understanding the question? or Are you having trouble understanding some of the terms involved? – TenaliRaman May 10 '12 at 14:18
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Let $S$ be a set, and let $x\in S$. Suppose that $x$ minimizes $f(x)$ over $S$. That means that $f(x)\le f(s)$ for all $s\in S$.
Suppose now that $R \subseteq S$. Then certainly $f(x)\le f(r)$ for all $r\in R$, because every $r\in R$ is also in $S$. So if $x\in R$, then $x$ minimizes $f$ over $R$.
The several mathematical symbols in the above argument tend to hide the simplicity of the logic. If Xavier ("$x$") is the shortest person in the United States ($S$), then Xavier is the shortest person in Rochester, NY ($R$).
André Nicolas
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