does the infimum of a finite, partially-ordered subset of R^2 ever not exist?

**wil** · February 24th 2010, 07:49 PM

I'm helping a friend with his homework, and the following problem is really giving us both trouble.

Consider the subset $B$ of $\mathbb{R}^{2}$ given by $B=\left[\left([0,1]\times[0,3]\right)\cup\left([0,2]\times[0,2]\right)\cup\left([0,3]\times[0,1]\right)\right]\backslash\left\{\left(2,1\right)\right\}$ . Give $B$ the partial order $\le'$ defined by $(a,b)\le'(c,d)$ if and only if $a\le c$ and $b\le d$ , where $\le$ is the usual order on $\mathbb{R}$ .

I'm good so far, but this is the part that has me stumped:

Find a set in $B$ which has no infimum.

It seems to me that any finite subset of $B$ would have an infimum.

(It should be noted that the book that this problem came from is well known for its typos and wrong answers in the student solutions section.)

**tonio** · February 24th 2010, 08:00 PM

Originally Posted by wil

I'm helping a friend with his homework, and the following problem is really giving us both trouble.

Consider the subset $B$ of $\mathbb{R}^{2}$ given by $B=\left[\left([0,1]\times[0,3]\right)\cup\left([0,2]\times[0,2]\right)\cup\left([0,3]\times[0,1]\right)\right]\backslash\left\{\left(2,1\right)\right\}$ . Give $B$ the partial order $\le'$ defined by $(a,b)\le'(c,d)$ if and only if $a\le c$ and $b\le d$ , where $\le$ is the usual order on $\mathbb{R}$ .

I'm good so far, but this is the part that has me stumped:

Find a set in $B$ which has no infimum.

It seems to me that any finite subset of $B$ would have an infimum.

(It should be noted that the book that this problem came from is well known for its typos and wrong answers in the student solutions section.)

.

**Drexel28** · February 24th 2010, 08:13 PM

Originally Posted by wil

I'm helping a friend with his homework, and the following problem is really giving us both trouble.

Consider the subset $B$ of $\mathbb{R}^{2}$ given by $B=\left[\left([0,1]\times[0,3]\right)\cup\left([0,2]\times[0,2]\right)\cup\left([0,3]\times[0,1]\right)\right]\backslash\left\{\left(2,1\right)\right\}$ . Give $B$ the partial order $\le'$ defined by $(a,b)\le'(c,d)$ if and only if $a\le c$ and $b\le d$ , where $\le$ is the usual order on $\mathbb{R}$ .

I'm good so far, but this is the part that has me stumped:

Find a set in $B$ which has no infimum.

It seems to me that any finite subset of $B$ would have an infimum.

(It should be noted that the book that this problem came from is well known for its typos and wrong answers in the student solutions section.)

Has an infimum in what sense? That the infimum exists in $B$ ?

**wil** · February 24th 2010, 10:11 PM

Originally Posted by Drexel28

Has an infimum in what sense? That the infimum exists in $B$ ?

You have a point there. (Hah! Get it? Point?) I guess if the problem is asking for a set that doesn't have an infimum in $B$ , then $[2,3]\times \{1\}$ would work.

**Drexel28** · February 24th 2010, 10:18 PM

Originally Posted by wil

You have a point there. (Hah! Get it? Point?) I guess if the problem is asking for a set that doesn't have an infimum in $B$ , then $[2,3]\times \{1\}$ would work.

I agree

**wil** · February 24th 2010, 10:21 PM

I'm thinking it might just be a question that could be worded more clearly. I'm pretty certain that every subset of $B$ has an infimum in $\mathbb{R}^2$ .

**Drexel28** · February 24th 2010, 10:26 PM

Originally Posted by wil

I'm thinking it might just be a question that could be worded more clearly. I'm pretty certain that every subset of $B$ has an infimum in $\mathbb{R}^2$ .

I guess what you're saying, and I need to think about it more is that does the LUB property of $\mathbb{R}$ transfer to $\mathbb{R}^2$ under the dictionary ordering.

**tonio** · February 25th 2010, 04:44 AM

Originally Posted by Drexel28

I guess what you're saying, and I need to think about it more is that does the LUB property of $\mathbb{R}$ transfer to $\mathbb{R}^2$ under the dictionary ordering.

This is the reason why I erased my first post: according to definition, if $S\subset P$ and $P$ is a part. ordered set , then an (in fact, THE) infimum of set is an element $z\in P$ (pay attention: in P!) s.t. the usual is fulfilled (as l.u.b. in the reals, say).

My guess is that the point $(2,1)$ that was taken out has something important to do here.

Tonio

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