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While I was studying metric spaces, I saw this question on the book:

Let $Q$ be the set of all rational numbers. What are $intQ$ and $∂Q$?

I think the answer should be: $intQ$ is all the rational numbers, and the $∂Q$ is all the real numbers. But I do not have the answer and do not feel confident with my answer. Is my answer true?

edit: To make it clear why I though $intQ$ is all the rational numbers: if I take $X_0$ = 3/2 and r=1, $(X_0-r,X_0+r)$ is a subset of $Q$ and as long as I choose $X_0$ a rational number, for all values of r, I can get a rational number.

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Why do you think that $(X_0-r,X_0+r)$ is a subset of $\mathbb Q$? –  TonyK Dec 3 '12 at 12:11
@TonyK Oh I guess I understood what you wanted to say. In the interval $(X_0-r,X_0+r)$, there are also irrational numbers and therefore it cannot be a subset of $Q$. Right? –  Amadeus Bachmann Dec 3 '12 at 12:14
Exactly.$$$$$$$$ –  TonyK Dec 3 '12 at 12:43
What if we wanted to find the closure and boundary of Q in R^2 instead of R? Does that make a difference? –  user51664 Dec 4 '12 at 1:21

2 Answers 2

up vote 3 down vote accepted

It depends on the topology we adopt. In the standard topology or $\mathbb{R}$ it is $\operatorname{int}\mathbb{Q}=\varnothing$ because there is no basic open set (open interval of the form $(a,b)$) inside $\mathbb{Q}$. Then, it follows that $\partial\mathbb{Q}=\mathbb{Q}$.

Comment on your edit: The definition of the internal of a set $C$ is the largest (wrt $\subseteq$) open set inside $C$. If you can't find a basic open set inside $C$ whatsoever, then the interior of $C$ is empty.

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How does it follow from $\operatorname{int}\mathbb{Q}=\varnothing$ that $\partial\mathbb{Q}=\mathbb{Q}$? –  Miikka Jan 25 at 17:51

You have to specify what metric space you are working in. If you say that $\mathbb{Q}$ is a subset of the metric space $(\mathbb{R},d)$ where $d$ is the usual metric given by the usual absolute value then we get that $\operatorname{int}{\mathbb{Q}} = \emptyset$ since if we take any $x \in \mathbb{Q}$ and $r \in \mathbb{R}$, $r>0$, then any open ball (i.e. open interval) centered at $x$ will contain irrationals, and hence it will not lie entirely in $\mathbb{Q}$.

Also we have $\partial \mathbb{Q} = \mathbb{R}$ since the boundary is defined to be the closure minus the interior, and the closure if simply $\mathbb{R}$ since if $x \in \mathbb{R}$ then $x$ ``adheres'' to $\mathbb{Q}$, i.e. there will be some sequence of rational numbers tending to $x$ (for instance, a decimal expansion).

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