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Suppose that $x^*$ satisfies $x^*\succsim x$ for $\forall x\in\{{x∈X|p·x\leq m}\}$.

How can we prove that $x\succsim x^*$ $\Rightarrow$ $p·x≥m$ if $\succsim$ is locally nonsatiated?

My idea for this goes like this but I am not sure:

Suppose that there exists $x' ∈ X$ such that $x'\succsim x^∗$ and $p · x' < m.$ Since we have local nonsatiation, for any $\epsilon>0$ there exists $x''\in X$ such that $||x''−x'||≤\epsilon$ and $x''\succ x'$. By transitivity of $\succsim$ and by $x''\succ x'\succsim x^*$. The former implies, by $p·x' < m$, that we can choose $\epsilon$ small enough such that $p·x'' < m$ and hence $x''\in\{{x∈X|p·x\leq m}\}$. This implies, by the definition of $x^∗$, that $x^∗\succsim x''$, a contradiction.

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    $\begingroup$ Your idea is largely correct, but there are a few typos. (1) In the 2nd sentence of the last paragraph, $x$ should have been $x'$; (2) where transitivity is invoked, the chain should have been $x''\succ x'\succsim x^*$; and (3) there's an extraneous "$\forall x \in$" in the second last sentence specifying the set to which $x''$ belongs. $\endgroup$
    – Herr K.
    Mar 1 at 19:31
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    $\begingroup$ I fixed them. I hope that it is correct now. Thank you for your input! $\endgroup$ Mar 1 at 19:42
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By local nonsatiation there exists a sequence $x_n\rightarrow x$ with $x_n\succ x\,\,\forall n$. By transitivity $x_n\succ x^*\,\,\forall n$. By contraposition $p\cdot x_n>m\,\,\forall n$. By continuity of the dot product $p\cdot x\ge m$.

This is pretty much a somewhat more compact version of your own approach.

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