I was aked to determine whether the following statement is true or false.

Let $$\mathcal{D} \left( (x_0, y_0), r \right) = \left\{ (x, y) \in \mathbb{R}^2 : (x - x_0)^2 + (y - y_0)^2 \leq r^2 \right\} $$ Let $P = \left\{ \emptyset \right\} \cup \left\{ \mathcal{D}\left( (x_0,y_0), r \right) : x_0, y_0 \in \mathbb{R}, r > 0 \right\} $. In the poset $(P, \subseteq ) $, there is always $\text{inf}\left\{ D_1, D_2\right\} $, for any $D_1, D_2 \in P$.

I think I came to the right conclusion but I'm thinking one could be more rigorous than I was. Here is what I attempted.

Consider that $\mathcal{D}\left( (x_0, y_0), r \right)$ is the set of points within a circumference with center $(x_0, y_0)$ and radius $r$. So we can think of $P$ as the set of all (filled) circles, including $\emptyset$.

Two circles may be related in one of the ways schematized by the following Venn diagrams:

enter image description here

Formally, for $D_1, D_2 \in P$, the image depicts the following exhaustive and mutually exclusive cases:

  • $D_1 \subseteq D_2$
  • $ D_1 \cap D_2 \neq \emptyset$ but $D_1 \not\subseteq D_2$
  • $D_1 \cap D_2 = \emptyset$.

(1) If $D_1 \subseteq D_2$ then evidently $\text{inf}\left\{ D_1, D_2 \right\} = D_1$, because $D_1$ would be the greatest circle that is a subset of itself and of $D_2$.

(2) If $D_1 \cap D_2 = \emptyset$, evidently $\text{inf}\left\{ D_1, D_2 \right\} = \emptyset$,.

(3) The case $D_1 \cap D_2 \neq \emptyset$ with $D_1 \not\subseteq D_2$ is the one that ruins the beauty. Let $D_3$ a circle s.t. $D_3 \subseteq D_1 \cap D_3$---this is, $D_3$ is an arbitrary, non-empty lower bound of $\left\{ D_1, D_2 \right\} $.

Observe that we can choose any arbitrary point $(z_1, z_2) \not\in D_3$ that lies in $D_1 \cap D_2$, and make $D_z = \mathcal{D}((z_1, z_2), \epsilon)$, with $\epsilon > 0$ a quantity sufficiently small to guarantee $D_z \cap D_3 = \emptyset$ and $D_z \in D_1 \cap D_2$. It is evident that $D_z \subseteq D_1$ and $D_z \subseteq D_2$, so $D_z$ is a lower bound of $\left\{ D_1, D_2 \right\} $; but since $D_z \not\subseteq D_3$ we cannot say $D_3$ is the greatest lower bound.

The argument above holds for any $D_3 \subseteq D_1 \cap D_2$. In general terms, we have shown that, in the case $D_1 \cap D_2 \neq \emptyset, D_1 \not\subseteq D_2$, for any lower bound $D_3$ of $\left\{ D_1, D_2 \right\} $, we can find a lower bound $D_z$ that is not a subset of $D_3$. Therefore no greater lower bound exists and there is no infimum.

Two questions arise. The most fundamental: Is this proof correct? But also, was there a simpler way to tackle the problem?

  • $\begingroup$ I think your proof is spot on! The appropriate level of rigour depends a bit on the broader context of the exercise - if you're a mathematician learning about posets, I think your proof looks fine. This type of "working" is a natural way to solve a problem like this, and it's valuable in & of itself to have a good understanding of exactly to what extent the property fails here. But do you understand that most of what you have written is not necessary in providing an answer to the problem? All you need is a single counterexample to answer the question, which you can do in a paragraph or so. $\endgroup$ Mar 4 at 0:06
  • 2
    $\begingroup$ If you want to pick some nits, "discs" would be the more widely used name of these objects, and I think your case 3 is really "the discs' interiors intersect, but neither is contained in the other", and 2 is "the interiors do not intersect" (since it's possible for the intersection to be a single point, in which case there is an infimum). $\endgroup$ Mar 4 at 0:11
  • 1
    $\begingroup$ @IzaakvanDongen Thanks, I'm happy to hear that. I am aware that all that I wrote is not necessary to answer the question; either a counter-example or simply describing the one case where the property does not hold would have sufficed. But I wanted to provide a general sense of how I approach the problem and what my understanding was of the whole matter, so if there were errors in my grasping people could point them out more easily. Thanks! $\endgroup$
    – lafinur
    Mar 4 at 0:25
  • $\begingroup$ Also I didn't knew the term "disk" referred to this type of sets, that's useful to know $\endgroup$
    – lafinur
    Mar 4 at 0:25
  • 2
    $\begingroup$ @dezdichado, infimum does have a fairly standard meaning in the context of partial orders - certainly it's not uncommon to talk about in the context of (complete) (semi-)lattices, for example. Also I think they're closed discs :-) $\endgroup$ Mar 4 at 15:09


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