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Is $S:=\{x\in\mathbb{Q}:x<\sqrt{2}\}$ closed in $\mathbb{Q}$ when endowed with the inherited topology from $\mathbb{R}$?

One argument for openness is that $S=\mathbb{Q}\cap (-\infty,\sqrt{2})$, which by definition of an open set in subspace topology, would force $S$ to be open.

On the other hand, $\mathbb{Q}\backslash S=\{x\in\mathbb{Q}:x\geq \sqrt{2}\}=\{x\in\mathbb{Q}:x>\sqrt{2}\}$, which by the preceding argument, should be open; ergo, $S$ is closed.

Which argument is correct? If both, $S$ is both open and closed which is unexpected. The context of this question is the determination of whether $\mathbb{Q}$ is connected as a topological space.

Sawyer
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    It is possible to have a set both open and closed. Both your arguments are correct. Infact, now you can say that $\mathbb{Q}$ is not connected. – Aniruddha Deshmukh Sep 11 '19 at 14:43
  • $S$ is open and closed hence so is its non empty complement. Hence $S$ is not connected. – copper.hat Sep 11 '19 at 14:43
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    By the way, I applaud you for asking an incredibly clear and well-expressed question, and showing that you'd done some thinking about it for yourself before asking it. Keep up the good work! – John Hughes Sep 11 '19 at 15:15

1 Answers1

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Yes, $S$ is both open and closed. Since $S$ is neither empty, nor equal to the whole space, this means that $\mathbb Q$ is not connected.

A somewhat more intuitive example that displays this same problem would be to consider the topological space $T=\mathbb R\setminus \{\sqrt{2}\}$ as a subspace of $\mathbb R$. We would expect this to be disconnected, as we cut a line at a point, giving a segment below the cut and one above. Indeed, the set $S=\{x\in T: x<\sqrt{2}\}$ is both open and closed in this space as well and $S$ and $T\setminus S$ turn out to be the connected components of $T$. This also makes sense with more typical ideas of what open sets are: if you are in $S$, then everything near enough to you is also in $S$, since you are some distance from the bound $\sqrt{2}$. This also applies to $T\setminus S$, as we remain a distance from $\sqrt{2}$.

The rationals are a subspace of this - so have only been cut apart even further, making the problem worse.

Milo Brandt
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