Let $\leq$ be the usual order on the field $\Bbb R$ of real numbers. Define an order $\leq$ on $\Bbb R^2$ by $(a,b) \leq (c,d)$ if $(a < c)$, or $(a = c$ and $b \leq d)$. Consider the subset $E$ = {${(\frac{1}{n} , 1 - \frac{1}{n} ) \in \Bbb R^2 : n \in \Bbb N}$}. With respect to $\leq$ which of the following statements is true ?
$\inf (E) = (0 ,1)$ and $\sup (E) = (0 , 1)$
$\inf (E)$ does not exist and $\sup (E) = (0 , 1)$
$\inf (E) = (0 ,1)$ and $\sup (E)$ does not exist
Both $\inf (E)$ and $\sup (E)$ do not exist.
We know that $\inf (E)$ = smallest member or smallest limit point of $E$ and $\sup (E)$ = largest point or largest limit point of $E$. By using above information, we get $\inf (E)$ does not exist and $\sup (E) = (0 , 1)$. I think, I'm not on right way. Please help me
You are correct about non-existence of infimum of $E$. I have tried to provide a proof as follows:
We see that $\forall n\in \mathbb{N}, a \in \mathbb{R}$ $$(0,a)<(\frac{1}{n}, 1-\frac{1}{n})$$ If $\exists (x,y)\in \mathbb{R^{2}}$ such that $$(0,a)<(x,y), \forall a\in \mathbb{R}$$ then $x>0$. Hence, using Archemedian Property, $\exists n_{0} \in \mathbb{N}$ such that $\frac{1}{n_{0}}<x$.
So we have $$(\frac{1}{n_{0}}, 1-\frac{1}{n_{0}})< (x,y).$$ Hence, if we had an infimum, it must be of the form $(0,a)$ which is not possible because $(0,a)<(0,a+1)$ and $(0,a+1)$ is again a lower bound of $E$.
So we don't have a "greatest" lower bound of $E$.
On the other hand the supremum of $E$ exists and $sup(E)=(1,0)$.
This is because $\forall n\in \mathbb{N}$, $$(\frac{1}{n}, 1-\frac{1}{n})\leq(1,0)$$ and $(1,0)\in E$.
So $(1,0)$ is the "least" upper bound of E.