For a set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show...
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I came across this question in a refrence book for discrete mathematics:
For a partially ordered set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show your answer for finitie and infinite set M.
After long thought about this question, I believe x is indeed the least element in case of finite set M. In case of infinite set M, x isn't the least element. I am unable to find the appropriate approach to prove this though.
Any help would be greatly appreciated!
discrete-mathematics
asked Nov 22 at 22:36
Harold J. Fike
83
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up vote
1
down vote
favorite
I came across this question in a refrence book for discrete mathematics:
For a partially ordered set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show your answer for finitie and infinite set M.
After long thought about this question, I believe x is indeed the least element in case of finite set M. In case of infinite set M, x isn't the least element. I am unable to find the appropriate approach to prove this though.
Any help would be greatly appreciated!
discrete-mathematics
asked Nov 22 at 22:36
Harold J. Fike
83
We have a partial order on the set, I guess. Do we assume anything more about this ordering?
– Berci
Nov 22 at 23:08
Yes, there is a partial order on the set. This is the only available piece of information. Forgot to mention, sorry!
– Harold J. Fike
Nov 22 at 23:16
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I came across this question in a refrence book for discrete mathematics:
For a partially ordered set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show your answer for finitie and infinite set M.
After long thought about this question, I believe x is indeed the least element in case of finite set M. In case of infinite set M, x isn't the least element. I am unable to find the appropriate approach to prove this though.
Any help would be greatly appreciated!
discrete-mathematics
asked Nov 22 at 22:36
Harold J. Fike
83
I came across this question in a refrence book for discrete mathematics:
For a partially ordered set M with x ∈ M being the only minimal element on that set, is x the least element as well? Show your answer for finitie and infinite set M.
After long thought about this question, I believe x is indeed the least element in case of finite set M. In case of infinite set M, x isn't the least element. I am unable to find the appropriate approach to prove this though.
Any help would be greatly appreciated!
discrete-mathematics
discrete-mathematics
asked Nov 22 at 22:36
Harold J. Fike
83
asked Nov 22 at 22:36
Harold J. Fike
83
edited Nov 22 at 23:17
asked Nov 22 at 22:36
Harold J. Fike
83
asked Nov 22 at 22:36
Harold J. Fike
83
asked Nov 22 at 22:36
Harold J. Fike
83
83
We have a partial order on the set, I guess. Do we assume anything more about this ordering?
– Berci
Nov 22 at 23:08
Yes, there is a partial order on the set. This is the only available piece of information. Forgot to mention, sorry!
– Harold J. Fike
Nov 22 at 23:16
add a comment |
We have a partial order on the set, I guess. Do we assume anything more about this ordering?
– Berci
Nov 22 at 23:08
Yes, there is a partial order on the set. This is the only available piece of information. Forgot to mention, sorry!
– Harold J. Fike
Nov 22 at 23:16
We have a partial order on the set, I guess. Do we assume anything more about this ordering?
– Berci
Nov 22 at 23:08
We have a partial order on the set, I guess. Do we assume anything more about this ordering?
– Berci
Nov 22 at 23:08
Yes, there is a partial order on the set. This is the only available piece of information. Forgot to mention, sorry!
– Harold J. Fike
Nov 22 at 23:16
Yes, there is a partial order on the set. This is the only available piece of information. Forgot to mention, sorry!
– Harold J. Fike
Nov 22 at 23:16
add a comment |
1 Answer
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In a finite set, every element is greater than or equal to some minimal element. Applying this to an arbitrary element $y$, shows that there is some minimal element below $y$. Since there is but one minimal element $x$, this proves that $xle y$. Hence $x$ is least.
In an infinite set, this property need not hold. Consider the set of integers $mathbb{Z}$, ordered in the usual way. It has neither minimal nor least elements. Let's also include one more element, $x$, that has the property that it incomparable to every integer. One can check that $x$ is minimal, but not least, in this partially ordered set $mathbb{Z}cup{x}$.
answered Nov 22 at 23:51
vadim123
75.2k896187
add a comment |
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1 Answer
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In a finite set, every element is greater than or equal to some minimal element. Applying this to an arbitrary element $y$, shows that there is some minimal element below $y$. Since there is but one minimal element $x$, this proves that $xle y$. Hence $x$ is least.
In an infinite set, this property need not hold. Consider the set of integers $mathbb{Z}$, ordered in the usual way. It has neither minimal nor least elements. Let's also include one more element, $x$, that has the property that it incomparable to every integer. One can check that $x$ is minimal, but not least, in this partially ordered set $mathbb{Z}cup{x}$.
answered Nov 22 at 23:51
vadim123
75.2k896187
add a comment |
up vote
0
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accepted
In a finite set, every element is greater than or equal to some minimal element. Applying this to an arbitrary element $y$, shows that there is some minimal element below $y$. Since there is but one minimal element $x$, this proves that $xle y$. Hence $x$ is least.
In an infinite set, this property need not hold. Consider the set of integers $mathbb{Z}$, ordered in the usual way. It has neither minimal nor least elements. Let's also include one more element, $x$, that has the property that it incomparable to every integer. One can check that $x$ is minimal, but not least, in this partially ordered set $mathbb{Z}cup{x}$.
answered Nov 22 at 23:51
vadim123
75.2k896187
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
In a finite set, every element is greater than or equal to some minimal element. Applying this to an arbitrary element $y$, shows that there is some minimal element below $y$. Since there is but one minimal element $x$, this proves that $xle y$. Hence $x$ is least.
In an infinite set, this property need not hold. Consider the set of integers $mathbb{Z}$, ordered in the usual way. It has neither minimal nor least elements. Let's also include one more element, $x$, that has the property that it incomparable to every integer. One can check that $x$ is minimal, but not least, in this partially ordered set $mathbb{Z}cup{x}$.
answered Nov 22 at 23:51
vadim123
75.2k896187
In a finite set, every element is greater than or equal to some minimal element. Applying this to an arbitrary element $y$, shows that there is some minimal element below $y$. Since there is but one minimal element $x$, this proves that $xle y$. Hence $x$ is least.
In an infinite set, this property need not hold. Consider the set of integers $mathbb{Z}$, ordered in the usual way. It has neither minimal nor least elements. Let's also include one more element, $x$, that has the property that it incomparable to every integer. One can check that $x$ is minimal, but not least, in this partially ordered set $mathbb{Z}cup{x}$.
answered Nov 22 at 23:51
vadim123
75.2k896187
answered Nov 22 at 23:51
vadim123
75.2k896187
answered Nov 22 at 23:51
vadim123
75.2k896187
answered Nov 22 at 23:51
vadim123
75.2k896187
75.2k896187
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We have a partial order on the set, I guess. Do we assume anything more about this ordering?
– Berci
Nov 22 at 23:08
Yes, there is a partial order on the set. This is the only available piece of information. Forgot to mention, sorry!
– Harold J. Fike
Nov 22 at 23:16