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take the set ## S := \{ 2^{-m} + 3^{-n} + 5^{-p} | m,n,p \in \mathbb{N} \} ##

claim [itex] inf(S) = 0 [/itex]

proof by contradiction:

suppose ## A > 0 ## where A is the largest lower bound for S

then ## 1/2^m + 1/3^n + 1/5^p \geq A ##

i.e. ## 2^m + 3^n + 5^p \geq A(2^m3^n5^p) ##

let ## X = 2^m + 3^n + 5^p ## and ## Y = 2^m3^n5^p ## where ## Y,X \in \mathbb{N} ##

then ## X \geq AY ## hence ## Y \leq \frac{X}{A} ## which is a contradiction to the archimedean postulate which states for any real x a natural number n >= x hence the lower bound is 0

is this proof OK or have I gone wrong somewhere?