The question was: What is the infimum from $K=\{x ~ |~0 \leq x \lt 1\}$ for the partial order $\{ (x,y) ~ | ~ ( \exists z \in \mathbb{R}_{\geq 0})[x-z=y] \}$ on set $\mathbb{R}$.
This was a question in a math quiz from our prof and the answer was $1$ but I don't get it why. I thought the infimum doesn't exist here. Can somebody explain it to me?
On
Looking at the partial order, $x$ is related to $y$, if there exists $z\geq 0$, such that $x-z=y$. We can demystify this relation by noting that this is equivalent to saying: $x$ is related to $y$, if $x\geq y$, where "$\geq$" denotes the usual order on the real numbers.
In short, $x$ is smaller than or equal to $y$ in the new order, if $x$ is greater than or equal to $y$ in the usual order.
The infimum of the set $K$ under the new order will thus be equal to the supremum of the set $K$ under the usual order, and this supremum is $1$.
Let $R$ be the partial order in question. For convenience write $x\preceq y$ if and only if $\langle x,y\rangle\in R$. Then $x\preceq y$ if and only if there is a $z\in\Bbb R_{\ge 0}$ such that $x-z=y$. But this is the case if and only if $x\ge y$, so for any $x,y\in\Bbb R$ we have $x\preceq y$ if and only if $x\ge y$. Thus, $\preceq$ is not just a partial order: it’s a linear (or total) order on $\Bbb R$. In fact, it’s simply the usual order $\ge$, the reverse of $\le$, so the infimum of a set with respect to $\preceq$ is the same as the supremum of the set with respect to $\le$. In the case of the set $K$, this is $1$.
If in doubt, you can check this directly against the definition. First, is $1$ a lower bound for $K$ with respect to $\preceq$? Yes: $1\ge x$ for each $x\in K$, so $1\preceq x$ for each $x\in K$. On the other hand, suppose that $b$ is a lower bound for $K$ with respect to $preceq$. Then $b\preceq x$ for each $x\in K$, so $b\ge x$ for each $x\in K$, and therefore $b\ge 1$. But then $b\preceq 1$, so $1$ is indeed the greatest lower bound for $K$ with respect to $\preceq$, i.e., its $\preceq$-infimum.