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**Opalg** What you did was more or less the right approach, but it has one serious error, right at the start, where you talk about "the smallest real number in T". There is no reason to suppose that T contains a smallest member. (For example, T might consist of all the strictly positive real numbers, with S consisting of all the negative real numbers together with 0.)

What you should do is to start by saying that T is bounded below (because S is nonempty, and every member of S is a lower bound for T). So T has an inf, call it $\displaystyle \beta$. *[That's the key point of the proof: T may not have a smallest member, but it does have an inf.]*

If x is any real number greater than $\displaystyle \beta$ then there exists a real number $\displaystyle t\in T$ with $\displaystyle \beta<t<x$ (otherwise x would be a lower bound for T, contradicting the fact that $\displaystyle \beta$ is the greatest lower bound for T). Therefore $\displaystyle x\notin S$ and hence $\displaystyle x\in T$.

On the other hand, if y is any real number less than $\displaystyle \beta$ then $\displaystyle y\notin T$ and hence $\displaystyle y\in S$.