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Consider: $A = \{x \in \mathbb R \mid \exists n \in \mathbb N : 2n-1 < x < 2n \}$.

How to find minimum/maximum/infimum/supremum (when exists)? I can't imagine this set in my head, so it's hard for me to find those values.

Regards

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$A$ is a set of real numbers $x$, that are strictly between an odd integer (2n - 1) and an even integer (2n). –  The Chaz 2.0 Nov 4 '11 at 12:35
    
@TheChaz: Note that integers could mean negative numbers as well. –  Asaf Karagila Nov 4 '11 at 12:37
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@Asaf: So what you're saying (and this comment is obviously not for your benefit) is that the definition of $A$ involves natural numbers, but I have (carelessly) included negative integers in my comment. Agreed. –  The Chaz 2.0 Nov 4 '11 at 15:48

2 Answers 2

I will assume that $0\in\mathbb N$ for the purpose of the question, if this is not the convention followed in your course, you should be able to adjust my calculations on your own.

If $x\in A$ then it means for some $n\in\mathbb N$ we have that $x\in (2n-1,2n)$. The least possible $n$ is $0$, so $x\in A$ would have to be greater than $-1$. However, note that $-1\notin A$ since it is not strictly smaller than $2\cdot 0-1$.

For the upper bound, given any real number $x$ there is some $k\in\mathbb N$ such that $x<2k$, so there exists some $y\in\mathbb R$ such that $x<y$ and $y\in A$. This should give you a hint about the boundedness of $A$.

Lastly, notice that: $$x\in\bigcup_{i\in I} A_i\iff\exists i\in I: x\in A_i$$

Set $I=\mathbb N$ and $A_n=(2n-1,2n)$ which is an open interval, and this will help you visualize the set.

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Hint:

If you take some values to $n$ is easy see the set:

$n=1 \implies A = \{x \in \mathbb R: 1 < x < 2 \}$

$n=2 \implies A = \{x \in \mathbb R: 3 < x < 4 \}$

,etc.

So if you get a generic $n$ value you are searching for the interval $(2n-1,2n)$.

Note too, min $2n-1$ happens when $n=0 (n \in N)$ and $2n-1=-1$.

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