Infimum of the set ${xin mathbb{Q};|;x^2<2}$?
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My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$
for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".
Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?
supremum-and-infimum
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
asked Dec 23 '18 at 9:31
MewMew
205
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add a comment |
$begingroup$
My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$
for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".
Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?
supremum-and-infimum
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
asked Dec 23 '18 at 9:31
MewMew
205
$endgroup$
2
$begingroup$
infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38
add a comment |
$begingroup$
My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$
for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".
Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?
supremum-and-infimum
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
asked Dec 23 '18 at 9:31
MewMew
205
$endgroup$
My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$
for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".
Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?
supremum-and-infimum
supremum-and-infimum
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
asked Dec 23 '18 at 9:31
MewMew
205
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
asked Dec 23 '18 at 9:31
MewMew
205
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
edited Dec 23 '18 at 10:25
Asaf Karagila♦
305k33436767
305k33436767
asked Dec 23 '18 at 9:31
MewMew
205
asked Dec 23 '18 at 9:31
MewMew
205
asked Dec 23 '18 at 9:31
MewMew
205
205
2
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infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38
add a comment |
2
$begingroup$
infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38
2
2
$begingroup$
infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38
$begingroup$
infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38
add a comment |
2 Answers
2
active
oldest
votes
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Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).
Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
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Would the first case not mean there is also no supremum, though?
$endgroup$
– Mew
Dec 23 '18 at 9:39
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Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
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– drhab
Dec 23 '18 at 9:41
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Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
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– Mew
Dec 23 '18 at 9:44
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I think so, yes.
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– drhab
Dec 23 '18 at 9:44
add a comment |
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To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).
Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
$endgroup$
$begingroup$
Would the first case not mean there is also no supremum, though?
$endgroup$
– Mew
Dec 23 '18 at 9:39
$begingroup$
Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
$endgroup$
– drhab
Dec 23 '18 at 9:41
$begingroup$
Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
$endgroup$
– Mew
Dec 23 '18 at 9:44
$begingroup$
I think so, yes.
$endgroup$
– drhab
Dec 23 '18 at 9:44
add a comment |
$begingroup$
Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).
Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
$endgroup$
$begingroup$
Would the first case not mean there is also no supremum, though?
$endgroup$
– Mew
Dec 23 '18 at 9:39
$begingroup$
Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
$endgroup$
– drhab
Dec 23 '18 at 9:41
$begingroup$
Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
$endgroup$
– Mew
Dec 23 '18 at 9:44
$begingroup$
I think so, yes.
$endgroup$
– drhab
Dec 23 '18 at 9:44
add a comment |
$begingroup$
Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).
Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
$endgroup$
Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).
Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
answered Dec 23 '18 at 9:38
drhabdrhab
102k545136
102k545136
$begingroup$
Would the first case not mean there is also no supremum, though?
$endgroup$
– Mew
Dec 23 '18 at 9:39
$begingroup$
Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
$endgroup$
– drhab
Dec 23 '18 at 9:41
$begingroup$
Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
$endgroup$
– Mew
Dec 23 '18 at 9:44
$begingroup$
I think so, yes.
$endgroup$
– drhab
Dec 23 '18 at 9:44
add a comment |
$begingroup$
Would the first case not mean there is also no supremum, though?
$endgroup$
– Mew
Dec 23 '18 at 9:39
$begingroup$
Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
$endgroup$
– drhab
Dec 23 '18 at 9:41
$begingroup$
Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
$endgroup$
– Mew
Dec 23 '18 at 9:44
$begingroup$
I think so, yes.
$endgroup$
– drhab
Dec 23 '18 at 9:44
$begingroup$
Would the first case not mean there is also no supremum, though?
$endgroup$
– Mew
Dec 23 '18 at 9:39
$begingroup$
Would the first case not mean there is also no supremum, though?
$endgroup$
– Mew
Dec 23 '18 at 9:39
$begingroup$
Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
$endgroup$
– drhab
Dec 23 '18 at 9:41
$begingroup$
Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
$endgroup$
– drhab
Dec 23 '18 at 9:41
$begingroup$
Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
$endgroup$
– Mew
Dec 23 '18 at 9:44
$begingroup$
Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
$endgroup$
– Mew
Dec 23 '18 at 9:44
$begingroup$
I think so, yes.
$endgroup$
– drhab
Dec 23 '18 at 9:44
$begingroup$
I think so, yes.
$endgroup$
– drhab
Dec 23 '18 at 9:44
add a comment |
$begingroup$
To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
$endgroup$
To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
answered Dec 23 '18 at 9:36
ArthurArthur
117k7116200
117k7116200
add a comment |
add a comment |
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$begingroup$
infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38