Proving that any lower bounded set has an infimum

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I've reached a point in a larger proof where I determined that a subset of the rational numbers, $X_n$ is bounded from below. I now need to prove that $X_n$ has an infimum in $\mathbb R$.

Intuitively this is clear, since the infimum would just be min($X_n$) if $X_n$ if it is a closed interval or isolated numbers, and would just be "$a$" if $X_n$ was an open interval ($a$,$b$). However, I am struggling to convert this into an actual proof.

My attempt was to just choose the lower bound L such that $\exists$ L s.t. $\forall$ x $\in$ $X_n$, L $\leq$ x & if $\exists$ L' s.t L'$\leq$ x $\forall$ x $\in$ $X_n$, then L'$\leq$ L, which would make L the infimum, but that seems as though I'm only defining an infimum and not showing that it actually exists.

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The statement in your title is equivalent to the axiom of completeness. If you have a definition of the real numbers, for instance as Dedekind cuts of the rationals, then you can prove it.