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I came across a theorem from Rudin's book that, the subsequential limits of a sequence $\{p_n\}$ in a metric space X form a closed subset of X. However, I also just learnt that $\mathbb{Q} \subset \mathbb{R}$ is not a closed subset of $\mathbb{R}$. And my confusion arises because the rationals and irrationals form a partition of the set of $\mathbb{R}$, which would imply that the irrational numbers cannot be an open nor a closed subset of $\mathbb{R}$.

If we let $\mathbb{R}$ to be our metric space X from the above Theorem, where all subsequential limits of a sequence of rational numbers form a subset of $\mathbb{R}$ which will be the set of irrationals, does this mean that the set of all irrational numbers should form a closed subset of $\mathbb{R}$? What am I missing?

Thanks.

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The set of irrationals (and rationals) in $\mathbb{R}$ is neither open nor closed. They're both not open because any neighborhood around a point contains both rational and irrational numbers. Since they are each other's complements, this means neither can be closed.

Your mistake is here: $"\textbf{where all subsequential limits of a sequence of rational numbers form a subset of } \mathbb{R}$ $\textbf{which will be the set of irrationals}$"

A subsequential limit of a sequence of rational numbers is not necessarily irrational. For instance, consider the sequence $0,0,..$

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  • $\begingroup$ I see what you mean, is my mistake that: all irrational numbers are a subsequential limit of rationals, but the converse, that all subsequential limits of rationals are irrational is not true? $\endgroup$ – T J. Kim Apr 11 '18 at 2:47
  • $\begingroup$ and since all subsequential limits of the rationals is not equal to the set of irrationals, there is no supposed contradiction $\endgroup$ – T J. Kim Apr 11 '18 at 2:49
  • $\begingroup$ @TJ.Kim correct $\endgroup$ – pwerth Apr 11 '18 at 2:51

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