I have a question on Theorem 2.41 on Page 40 of Walter Rudin's Principles of Mathematical Analysis, 3rd Edition. Have stated below the theorem, followed by my question.
Theorem 2.41: If a set $E$ in $R^k$ has one of the following three properties, then it has the other two:
(a) $E$ is closed and bounded,
(b) $E$ is compact, and,
(c) every infinite subset of $E$ has a limit point in $E$.
My question: Is the following situation a counterexample to the assertion that (c) $\implies$ (a)? Let $E$ be the set of all rational numbers in $R^1$. It appears (to me) that every infinite subset of $E$ has a limit point in $E$. Thus, (c) holds in my understanding. However, $E$ is not closed (since $E$ does not include all its limit points), and $E$ is not bounded either. This means that (a) does not hold.
I also see that (similar?) questions have been raised earlier on math.stackexchange.com on the above-mentioned theorem, and been answered. But, no question has cited the above-mentioned "counterexample" (or similar) in my observation. Would greatly appreciate help in understanding where this "counterexample" is wrong (which should be the case in all likelihood I believe). Thanks.
Best Answer
For every irrational $x$, there is a sequence of rationals that converges to $x$ (and hence has no rational limit point). E.g. you could use the truncated decimal expansion.