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Had some trouble with this question in an exam recently, and wanted to make sure I reasoned correctly. The question was:

$X$ is a set of pairs of real numbers $(x,y)$, with absolute values less than or equal to $10$.

$R$ is an order relation on $X$ defined as $$ R=\lbrace \space (x,y), (x',y') \in X \times X \space |\space x \leq x' \wedge y \leq y' \space \rbrace $$ with $\leq$ the usual order on the reals.

Find the meet (greatest lower bound) and join (greatest upper bound) of the subset $A$, where $$ A=\lbrace \space (x,y) \in X \times X \space | \space (x-1)^2 + (y-2)^2 = 100 \space \rbrace $$

Now am I right in saying that while there are a number of pairs $(x,y)$ that will satisfy $A$, these are not necessarily comprable and that therefore no glb or lub exists for $A$?

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Consider the points $(10, 2 + \sqrt{19})$ and $(7, 10)$, which are in $A$. Any $(x, y)$ with $x < 10$ or $y < 10$ will not be greater than these points. Moreover, $(10, 10)$ is an upper bound for all of $X$, and $A \subset X$. This proves that $(10, 10)$ is the least upper bound for $A$.

Likewise, if $(x, y) \in X$ such that $x < -9$ or $y < -8$, then $(x, y) \notin A$. Hence $(-9, -8)$ is a lower bound for $A$. Since $(-9, 0), (0, -8) \in A$, it is in fact the greatest lower bound.

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  • $\begingroup$ Right, so I was specifically looking for a least and greatest pair that at the same time satisfy the condition that (x-1)^2 + (y-2)^2 = 100, where I should have been looking for boundaries on the whole of A. This helps lots, thank you. $\endgroup$ – Charl Cater Jun 10 '14 at 15:57
  • $\begingroup$ @CharlCater Yeah, that was my confusion on the comment I posted originally (now deleted). Just because $A$ doesn't have a lub which is in $A$ doesn't mean no lub exists. $\endgroup$ – Alex G. Jun 11 '14 at 3:27
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No, this is wrong. Take your set, which is bounded, and enclose it as tightly as possible into a rectangle with sides parallel to $x$, $y$ axes. Then the greatest lower bound is the lower left corner, and lowest upper bound is the upper right corner of the rectangle.

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  • $\begingroup$ OK great, that makes sense. Thank you for clarifying. $\endgroup$ – Charl Cater Jun 10 '14 at 15:25

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