What is $sup{ (-1)^n + 1/n} $?

Question

My doubt is that $\sup (A+B) = \sup A + \sup B$. So, if we consider $A =(-1)^n$ and $B=\frac{1}{n},$

then $\sup A = 1 $ and $\sup B = 1$. So will $\sup (A+B) = \sup{ (-1)^n + 1/n} = \sup (-1)^n + \sup \frac{1}{n} = 1+1 =2 ?$

Is this correct or have I misunderstood somewhere please explain?

Also could you please provide the correct method to solve the above problem?

See e.g. How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B={a+b\mid a\in A, b\in B}$ for proof of the statement $\sup(A+B)=\sup(A)+\sup(B)$. — Minus One-Twelfth, May 30 '19 at 12:57
Yes, I did. You said you had doubt about $\sup(A+B)=\sup(A)+\sup(B)$, so I provided a link about that. — Minus One-Twelfth, May 30 '19 at 13:00
@MinusOne-Twelfth I don't need the proof , what I need is the clarification of the e.g I gave — user678450, May 30 '19 at 13:00
So you don't actually have doubt that $\sup(A+B)=\sup A + \sup B$? I was just providing that link to address that doubt, not as an answer to your question. — Minus One-Twelfth, May 30 '19 at 13:01
@MinusOne-Twelfth yeah I have no doubt regarding that . Thanks for the help — user678450, May 30 '19 at 13:02
$\newcommand{\Z}{: n\in \mathbb{Z}^{+}}$By the way, that theorem isn't applicable like that here since ${(-1)^n + 1/n \Z } \ne { (-1)^n \Z } + { 1/n \Z }$. I.e. the set you need the sup of is not actually equal to $A+B$. (Where $A = { (-1)^n \Z }$ and $B = {1/n \Z }$.) — Minus One-Twelfth, May 30 '19 at 13:03
@MinusOne-Twelfth why? Could you please explain that , this was my real problem — user678450, May 30 '19 at 13:04

score 4 · Answer 1 · answered May 30 '19 at 13:03

The problem here is that in the statement $$ \sup (A + B) = \sup A + \sup B $$ the definition of "$+$" is given by $$ A+B=\{a+b\mid a\in A, b\in B\}, $$ and $A$ and $B$ are sets (of real numbers) .

You, in trying to apply it, appear to be adding two sequences termwise, i.e., you're defining a new sequence $$ (A+B)_n = A_n + B_n $$ With this definition of addition (and the implied notion that "sup" of a sequence is the sup of the set of items in the sequence) the "theorem" is no longer a theorem, as your counterexample shows.

score 2 · Answer 2 · answered May 30 '19 at 13:07

You cannot use this property. This is because were are not talking about elements in sets, but about sequences which depend on the same $n$, therefore, these sequences are not 'disjoined'.

An easy way to look at this problem is that you want to "maximize" both $(-1)^n$ and $1/n$ simultaneously. Notice that both are bounded by $1$, so we want to show that we can find $n \in \mathbb{N}$ such that both components are equal to one. Notice that for $(-1)^n=1$, we must have that $n$ is even, i.e. $n=2k$. Substituting the later into the second component, we have $\frac{1}{n} = \frac{1}{2k}$, where $\sup\big({\frac{1}{2k}}\big) = \frac{1}{2}$, where $k=1$, and thus $n = 2k = 2$. Hence, \begin{equation} \sup\bigg(\frac{1}{n} + (-1)^n\bigg) = \frac{1}{2} + 1 = 1\frac{1}{2} \end{equation}

yeah, I have found my mistake – user678450 May 30 '19 at 13:17 — user678450, May 30 '19 at 13:17

score 1 · Accepted Answer · answered May 30 '19 at 13:11

$\newcommand{\s}{: n\in\mathbb{Z}^{+}}$ If $A = \{ (-1)^n \s \}$ and $B = \{ 1/n \s \}$, it is not true that $\{ (-1)^n + 1/n \s \}$ equals $A+B$. This is because $A+B$ refers to the set of all sums of pairs of elements of $A$ and $B$ where the "$n$" can be different for each. That is, $A+B = \{ (-1)^{n_1} + 1/n_2 : n_1, n_2\in \mathbb{Z}^{+}\}$, whereas in $\{ (-1)^n + 1/n \s \}$, the "$n$" is the same in both terms.

For example, $(-1)^3 + 1/2 = -1/2$ is in $A+B$ but you can show that $-1/2$ is not in $\{ (-1)^n + 1/n \s \}$.

Thus, your reasoning was unfortunately incorrect.

(Written using functions, you can say that $\sup\limits_n (f(n)+g(n))$ in general does not equal $\sup\limits_n f(n)+\sup\limits_n g(n)$.)

What is $sup{ (-1)^n + 1/n} $?

3 Answers3