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Suppose ## S \subset \mathbb{R} ## is a nonempty subset of the real numbers that is bounded below.

Show that there exists a sequence ## <x_n> ## such that ## x_n \in S ## for all n and ## \lim (x_n) = inf(S) ##

attempt:

consider an element ## x \in S ## suppose ## x \geq inf(S) + 1/n ## then this would mean that inf(S) is not the greatest lower bound and inf(s) + 1/n is so ## x < inf S + 1/n ## ## \forall n \in \mathbb{N} ## take ## n_1 < n_2 < n_3 < n_4 ... ## then ## x_1 < inf(S) + 1/n_{1} ## and ## x_2 < inf(S) + 1/n_{2} ## and so on so we get a sequence ## inf(S) \leq x_n ... < x_3 < x_2 < x_1 ## hence as n increases 1/n approaches zero (proved already) and x_n approaches inf(S).

Is this all correct as in the solutions all they have stated is:

## inf(S) \leq x_n < inf(S) + 1/n ## and inf(s) + 1/n -> inf(S) so such a sequence exists, I don't understand that