Solution for Exercise 04 from Chapter 01 in Baby Rudin

[gapry]

This pull request provides the solution for Exercise 04, Chapter 01, from the third edition of Principles of Mathematical Analysis by Walter Rudin.

[ceciliachan1979]

When you write $\alpha \in S$ in the definition of $L$, you redefined $\alpha$ in that context, but $\alpha$ is already defined by the problem. Try not to redefine symbol, that can get confusing, same goes with $\beta$.

[ceciliachan1979]

You defined $L$ and $U$, but you never used it.

[ceciliachan1979]

The whole thing could be simplify to just this, you don't really need to argue the inequalities for all elements in the set in order to claim the result.

Let $S$ is an ordered set, and let $E$ be a non-empty subset of $S$, so there exists $x \in E$.

$\alpha$ is a lower bound of $E$, so $\alpha \le x$,

$\beta$ is an upper bound of $E$, so $x \le \beta$,

Combined, we have $\alpha \le x \le \beta$, so $\alpha \le \beta$