Infimum and supremum of the set $\{(\frac{1}{n},1-\frac{1}{n}):n\in \mathbb{N}\} $ with dictionary order in $\mathbb{R}^2$ Geometrically thinking, it goes from $(1,0)$ and approaches $(0,1)$.
So, my guess is infimum is $(0,1)$ and supremum is $(1,0)$.
Is it true? can someone help?
 A: for any $n\in \mathbb N$ we have $\frac 1n > 0$ so any $(0,x) < (\frac 1n,1-\frac 1n)$ so any $(0, x)$ is lower bound. 
If $k > 0$ then there is an $n \in \mathbb N$ so that $0< \frac 1n < k$ so $(\frac 1n,1-\frac 1n) < (k,x)$ so no $(k,x);k \ge 0$ can be and lower bound.
(And it pretty much goes without saying, but for the sake of being thorough, if $k < 0$ then $(k,x) < (0,y)$ for all $x,y$. So, although all $(k,x)$ are all lower bounds to our set, none of them are greatest lower bounds, as $(0,y)$ are all lower bounds that are higher.)
So if any $\inf$ of the set exists is of the form $(0,x)$.  But for every $(0,x)$ that is a lower bound there is a $y>x$ so that $(0,y)> (0,x)$ but $(0,y)$ is also a lower bound.  So there is no $\inf$ of the set.
.....
If $n \in \mathbb N$ then either $n =1$ and $(\frac 1n, 1-\frac 1n) = (1,0)$ or $n > 1$ and $(\frac 1n, 1-\frac 1n) < (1,0)$.  So $(1,0) \ge (\frac 1n, 1-\frac 1n)$ for all elements in the set.  So $(1,0)$ is an upper bound.
But $(1,0) = (\frac 11, 1-\frac 11)$ is in the set. 
So if $(x,y) < (1,0)$ then $(x,y)$ can not be an upper bound (because $(1,0)$ is an element that is larger than $(x,y)$.).  So $(1,0)$ is the $\sup$ of the set..
===
Addendum:  Don't forget the actual definition and a very basic test for $\inf$ and $\sup$ is that the $\sup$ ($\inf$) is a upper (lower) bound so that any element that is less (greater) than it is not an upper (lower) bound.
So as every $(0,x)$ does have a $(0, y) $ greater than it that is a lower bound, no $(0,x)$ can be an $\inf$.  And as nothing lower than $(1,0)$ can be an upper bound while $(1,0)$ then $(1,0)$ is $\sup$.
A: Any point $(0,x)$ is a lower bound, and to be a lower bound for the set, the first coordinate has to be $\le 0$ (if $(a,b)$ with $a>0$ then for large $n$, $\frac1n < a$ and $(\frac1n, 1-\frac1n) <  (a,b)$ and so no lower bound).
So there is no greatest lower bound (aka infimum) ($(0,b)$ is always smaller than $(0,b+1)$ etc.)
As to upperbounds, $(1,0)$ is the maximum of the set (take $n=1$) and thus it’s supremum too. 
