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A set $A = \left\{(-1)^n + \frac{1}{m} : n,m \in \Bbb N\right\} \cup \{-1\}$ is given.

a) I shall find out and justify what the supremum an infimum is.

b) Is the sup a maximum or the inf a minimum.

a) For the sup, I find out $n=2$ and $m=1$ because then I have $1+1=2$. For the inf, I have $\displaystyle \lim _{m\to\infty} (1/m) = 0$ and $n = \text{odd number} \implies 0-1=-1$.

Now my questions are: Is that correct? How can I now find out if there is a max or min? Can I say for the sup that $2$ and $1$ are elements of $\Bbb N$ so the sup is a max and for the inf same? And what about the $\{-1\}$ in the set, what does this $\{-1\}$ mean for the inf, sup, min and max?

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    For a very similar problem(and solution) see [here](http://math.stackexchange.com/questions/810064/supremum-of-two-subtracted-fractions-less-than-one/810095#810095).2014-06-04

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Every element $x \in A$ is $x\leq2$(why?) Thus $2$ is an upper bound of $A$. Can you find a lower upper bound of $A$? The smallest of the upper bounds is the $\sup A$.

Every element $x \in A$ is $x\geq-1$(why?) Thus $-1$ is a lower bound of $A$. Can you find a larger lower bound of $A$? The largest of the lower bounds is the $\inf A$.

Edit: You got it right $\sup A=2$ and $\inf A=-1.$
To prove this you must show that $2$ is an upper bound of $A$ (i.e. $x\leq 2, \ \forall x \in A$ ) and that for any $\epsilon>0, \ 2-\epsilon$ is not an upper bound of $A$(i.e. there is an $x \in A$ with $x>2-\epsilon$) . SImilar with $\inf A$.

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    And for $x\leq4$ I would assume 4 is an upper bound because it is the bigger than 2, because 2 is the heighest result I can get.Ok so there is no maximum because the max would be $+\infty$, because $n,m \in N$ .Ah now I also believe that I have understood {-1} is mentionned,it is because of {N^+} ... as I see Patrick have mentionned that already.2012-11-29
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It is very easy. Write A = {−1 + 1/m :m∈N} U {1 + 1/m :m∈N}. (Prove this equality) Then sup A = max{supB , supC} where B = {−1 + 1/m :m∈N} and C= {1 + 1/m :m∈N} (also prove it very easy) now B = {-1} + {1/m :m∈N}. where addition of set means def of B. Now sup{1/m :m∈N} = 0(by archimedean property) and sup B = sup{-1} +sup{1/m :m∈N} (prove this one also) similar find for C and you will get. sup A = 2 and inf A = -1 But there is no maximum and minimum value. Because max and min value if exist are equal to sup and inf respectively(easy to prove this statement) And it is easy to show 2 and -1 does not belong to A.

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For part a) begin by thinking of the $(-1)^n$ and the $\frac{1}{m}$ parts as individual sets and what their supremums and infimums are. First one just oscillates between $-1$ and $1$ for example. The $\{-1\}$ part is just throwing $-1$ into the set basically. You should be able to see that you cannot get $-1$ as the sum of $(-1)^n + \frac{1}{m}$ using the domain given. (also I think you meant to write $n,m \in \mathbb{N^+}$.

For part b) the infimum is the minimum if it is actually an element of the set and likewise for the supremum.

Hope that helps.

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    yes, you are right I wannted to write n=2.2012-11-29