Infimum and supremum of the set $ \{ (1)^n + 1/m : n,m \in \mathbb N \} \cup \{1\}$
Solution 1
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 $x\geq1$(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 $\inf A$.
Solution 2
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.
Related videos on Youtube
phil
Updated on August 01, 2022Comments

phil over 1 year
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 $n = \text{odd number} \implies 01=1$.
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?

hrkrshnn over 9 yearsFor a very similar problem(and solution) see here.


Patrick almost 11 years@ Phil Also check your math that when $n = m = 1$ we have $(1)^n + \frac{1}{m} = 2$

phil almost 11 yearsyes, you are right I wannted to write n=2.

phil almost 11 yearsSo it isn't sup(A)=2 and inf(A)=1 ?

phil almost 11 yearsAnd 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.