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Describe the interior, closure and boundary of the following sets in the real line:

  • the set of all integers

  • the set of all rationals

  • the set of all irrationals

  • $(0,1)$

  • $[0,1]$

Could someone help me through this problem?

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    Why don't you start with ONE of those sets, and try to find (say) the *closure* of that set. Write out the definition. See if you can't get some intuition from the wikipedia page and other online (or hard copy) resources.2012-04-22

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Let's do the integers together and then you can try the other ones on your own:

(i) $\operatorname{int}{\mathbb Z}$:

By definition the interior of a set $S$ are all the points $x$ in it such that we can find an $\varepsilon > 0$ such that $B(x,\varepsilon) \subset S$.

Let $\varepsilon > 0$ and $n \in \mathbb Z$. Our $\varepsilon$ is small but we know that if we pick a $k \in \mathbb N$ large enough then we can achieve \frac{1}{k} < \varepsilon. But then $n + \frac{1}{k} \in B(n, \varepsilon)$ and $n + \frac{1}{k} \notin \mathbb Z$, so $B(n,\varepsilon) \not\subset \mathbb Z$. So the interior of $\mathbb Z$ must be empty.

(ii) $\overline{\mathbb Z} \subset \mathbb R$:

We know that $\overline{\mathbb Z} = \partial \mathbb Z \cup \operatorname{int}{\mathbb Z}$ so let's find out what the boundary $\partial \mathbb Z$ is.

(iii) $\partial \mathbb Z$:

By definition, $\partial \mathbb Z$ are all points $n$ such that every $B(n, \varepsilon)$ has non-empty intersection with both, $\mathbb Z$ and $\mathbb R \setminus \mathbb Z$. Let $n$ be any point in $\mathbb Z$ and $B(n, \varepsilon)$ an epsilon ball around it. Then the intersection $\mathbb Z \cap B(n, \varepsilon)$ contains $n$ hence is non-empty. Also, in (i) we saw that every such epsilon ball has non-empty intersection with $\mathbb R \setminus \mathbb Z$. So $\partial \mathbb Z = \mathbb Z$.

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    @MartinSleziak Thank you, Martin, yes I meant subset rather than in, of course : ) It's early here and I've only had one cup of coffee so far : )2012-04-22