A sequence with an infimum and no minimum [closed]
If a sequence has an infimum but no minimum, does this mean the infimum is the limit of the sequence? If so how do you prove this?
analysis supremum-and-infimum
asked Dec 3 '18 at 20:34
Laura
112
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Alexander Gruber♦ Dec 4 '18 at 3:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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If a sequence has an infimum but no minimum, does this mean the infimum is the limit of the sequence? If so how do you prove this?
analysis supremum-and-infimum
asked Dec 3 '18 at 20:34
Laura
112
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Alexander Gruber♦ Dec 4 '18 at 3:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
3
The infimum is just the liminf, not necessarily limit.
– GNUSupporter 8964民主女神 地下教會
Dec 3 '18 at 20:38
4
The sequence ${a_n}$ defined by $$a_n = begin{cases} frac{1}{n} & text{$n$ is even,} \ 1 & text{otherwise}end{cases} $$ has infimum equal to zero, but has no limit.
– Xander Henderson
Dec 3 '18 at 20:44
2
@GNUSupporter8964民主女神地下教會 $$liminf a_n=inf a_n iff {a_n}text{ is monotonically decreasing }$$ Otherwise we always have $$ liminf a_nleinf a_n$$
– Daniele Tampieri
Dec 3 '18 at 20:55
1
@DanieleTampieri I left my previous comment too quickly. What about changing Xander Handerson's example from 1 to -1, in this way, liminf and inf are equal to -1, but $(a_n)$ is oscillating.
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 3:06
1
@GNUSupporter8964民主女神地下教會 I think it would be a nice counterexample. And also let me say that also I left my comment to quickly: a necessary premise for it would be "If ${a_n}$ is a converging sequence then" and the sign in the inequality between $liminf a_n$ and $inf a_n$ should be "$ge$", not "$le$".
– Daniele Tampieri
Dec 4 '18 at 5:59
|
show 1 more comment
If a sequence has an infimum but no minimum, does this mean the infimum is the limit of the sequence? If so how do you prove this?
analysis supremum-and-infimum
asked Dec 3 '18 at 20:34
Laura
112
If a sequence has an infimum but no minimum, does this mean the infimum is the limit of the sequence? If so how do you prove this?
analysis supremum-and-infimum
analysis supremum-and-infimum
asked Dec 3 '18 at 20:34
Laura
112
asked Dec 3 '18 at 20:34
Laura
112
asked Dec 3 '18 at 20:34
Laura
112
asked Dec 3 '18 at 20:34
Laura
112
asked Dec 3 '18 at 20:34
Laura
112
112
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Alexander Gruber♦ Dec 4 '18 at 3:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Alexander Gruber♦ Dec 4 '18 at 3:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
3
The infimum is just the liminf, not necessarily limit.
– GNUSupporter 8964民主女神 地下教會
Dec 3 '18 at 20:38
4
The sequence ${a_n}$ defined by $$a_n = begin{cases} frac{1}{n} & text{$n$ is even,} \ 1 & text{otherwise}end{cases} $$ has infimum equal to zero, but has no limit.
– Xander Henderson
Dec 3 '18 at 20:44
2
@GNUSupporter8964民主女神地下教會 $$liminf a_n=inf a_n iff {a_n}text{ is monotonically decreasing }$$ Otherwise we always have $$ liminf a_nleinf a_n$$
– Daniele Tampieri
Dec 3 '18 at 20:55
1
@DanieleTampieri I left my previous comment too quickly. What about changing Xander Handerson's example from 1 to -1, in this way, liminf and inf are equal to -1, but $(a_n)$ is oscillating.
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 3:06
1
@GNUSupporter8964民主女神地下教會 I think it would be a nice counterexample. And also let me say that also I left my comment to quickly: a necessary premise for it would be "If ${a_n}$ is a converging sequence then" and the sign in the inequality between $liminf a_n$ and $inf a_n$ should be "$ge$", not "$le$".
– Daniele Tampieri
Dec 4 '18 at 5:59
|
show 1 more comment
3
The infimum is just the liminf, not necessarily limit.
– GNUSupporter 8964民主女神 地下教會
Dec 3 '18 at 20:38
4
The sequence ${a_n}$ defined by $$a_n = begin{cases} frac{1}{n} & text{$n$ is even,} \ 1 & text{otherwise}end{cases} $$ has infimum equal to zero, but has no limit.
– Xander Henderson
Dec 3 '18 at 20:44
2
@GNUSupporter8964民主女神地下教會 $$liminf a_n=inf a_n iff {a_n}text{ is monotonically decreasing }$$ Otherwise we always have $$ liminf a_nleinf a_n$$
– Daniele Tampieri
Dec 3 '18 at 20:55
1
@DanieleTampieri I left my previous comment too quickly. What about changing Xander Handerson's example from 1 to -1, in this way, liminf and inf are equal to -1, but $(a_n)$ is oscillating.
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 3:06
1
@GNUSupporter8964民主女神地下教會 I think it would be a nice counterexample. And also let me say that also I left my comment to quickly: a necessary premise for it would be "If ${a_n}$ is a converging sequence then" and the sign in the inequality between $liminf a_n$ and $inf a_n$ should be "$ge$", not "$le$".
– Daniele Tampieri
Dec 4 '18 at 5:59
3
3
The infimum is just the liminf, not necessarily limit.
– GNUSupporter 8964民主女神 地下教會
Dec 3 '18 at 20:38
The infimum is just the liminf, not necessarily limit.
– GNUSupporter 8964民主女神 地下教會
Dec 3 '18 at 20:38
4
4
The sequence ${a_n}$ defined by $$a_n = begin{cases} frac{1}{n} & text{$n$ is even,} \ 1 & text{otherwise}end{cases} $$ has infimum equal to zero, but has no limit.
– Xander Henderson
Dec 3 '18 at 20:44
The sequence ${a_n}$ defined by $$a_n = begin{cases} frac{1}{n} & text{$n$ is even,} \ 1 & text{otherwise}end{cases} $$ has infimum equal to zero, but has no limit.
– Xander Henderson
Dec 3 '18 at 20:44
2
2
@GNUSupporter8964民主女神地下教會 $$liminf a_n=inf a_n iff {a_n}text{ is monotonically decreasing }$$ Otherwise we always have $$ liminf a_nleinf a_n$$
– Daniele Tampieri
Dec 3 '18 at 20:55
@GNUSupporter8964民主女神地下教會 $$liminf a_n=inf a_n iff {a_n}text{ is monotonically decreasing }$$ Otherwise we always have $$ liminf a_nleinf a_n$$
– Daniele Tampieri
Dec 3 '18 at 20:55
1
1
@DanieleTampieri I left my previous comment too quickly. What about changing Xander Handerson's example from 1 to -1, in this way, liminf and inf are equal to -1, but $(a_n)$ is oscillating.
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 3:06
@DanieleTampieri I left my previous comment too quickly. What about changing Xander Handerson's example from 1 to -1, in this way, liminf and inf are equal to -1, but $(a_n)$ is oscillating.
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 3:06
1
1
@GNUSupporter8964民主女神地下教會 I think it would be a nice counterexample. And also let me say that also I left my comment to quickly: a necessary premise for it would be "If ${a_n}$ is a converging sequence then" and the sign in the inequality between $liminf a_n$ and $inf a_n$ should be "$ge$", not "$le$".
– Daniele Tampieri
Dec 4 '18 at 5:59
@GNUSupporter8964民主女神地下教會 I think it would be a nice counterexample. And also let me say that also I left my comment to quickly: a necessary premise for it would be "If ${a_n}$ is a converging sequence then" and the sign in the inequality between $liminf a_n$ and $inf a_n$ should be "$ge$", not "$le$".
– Daniele Tampieri
Dec 4 '18 at 5:59
|
show 1 more comment
2 Answers
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As pointed out in the comments by @Xander Henderson, there are sequences that do not converge which have no minimum. This is because the infimum refers to the greatest lower bound of the sequence considered as a set in $mathbb{R}$. If the sequence $S={a_n}$ has an infimum but no minimum, this tells us there is a limit point of $S$ that is not contained in $S$. All we can say is that there is a subsequence ${a_{n_k}}$ that converges to the infimum.
The more useful notion you may be looking for is the lim inf, which is the infimum of all possible subsequential limits. Note that in general
$$inf_{n} a_n le liminf_{n to infty} a_n$$
since the infimum is a subsequential limit. Whenever a sequence converges, it indeed converges to the lim inf.
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
add a comment |
You can just create a series $a_n = left(frac{1}{n}right)$ defined on the real interval $(0,1]$. This way, the infimum is $0$, but there is no minimum because the interval is half-open.
answered Dec 3 '18 at 20:53
Thomas Lang
1624
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
As pointed out in the comments by @Xander Henderson, there are sequences that do not converge which have no minimum. This is because the infimum refers to the greatest lower bound of the sequence considered as a set in $mathbb{R}$. If the sequence $S={a_n}$ has an infimum but no minimum, this tells us there is a limit point of $S$ that is not contained in $S$. All we can say is that there is a subsequence ${a_{n_k}}$ that converges to the infimum.
The more useful notion you may be looking for is the lim inf, which is the infimum of all possible subsequential limits. Note that in general
$$inf_{n} a_n le liminf_{n to infty} a_n$$
since the infimum is a subsequential limit. Whenever a sequence converges, it indeed converges to the lim inf.
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
add a comment |
As pointed out in the comments by @Xander Henderson, there are sequences that do not converge which have no minimum. This is because the infimum refers to the greatest lower bound of the sequence considered as a set in $mathbb{R}$. If the sequence $S={a_n}$ has an infimum but no minimum, this tells us there is a limit point of $S$ that is not contained in $S$. All we can say is that there is a subsequence ${a_{n_k}}$ that converges to the infimum.
The more useful notion you may be looking for is the lim inf, which is the infimum of all possible subsequential limits. Note that in general
$$inf_{n} a_n le liminf_{n to infty} a_n$$
since the infimum is a subsequential limit. Whenever a sequence converges, it indeed converges to the lim inf.
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
add a comment |
As pointed out in the comments by @Xander Henderson, there are sequences that do not converge which have no minimum. This is because the infimum refers to the greatest lower bound of the sequence considered as a set in $mathbb{R}$. If the sequence $S={a_n}$ has an infimum but no minimum, this tells us there is a limit point of $S$ that is not contained in $S$. All we can say is that there is a subsequence ${a_{n_k}}$ that converges to the infimum.
The more useful notion you may be looking for is the lim inf, which is the infimum of all possible subsequential limits. Note that in general
$$inf_{n} a_n le liminf_{n to infty} a_n$$
since the infimum is a subsequential limit. Whenever a sequence converges, it indeed converges to the lim inf.
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
As pointed out in the comments by @Xander Henderson, there are sequences that do not converge which have no minimum. This is because the infimum refers to the greatest lower bound of the sequence considered as a set in $mathbb{R}$. If the sequence $S={a_n}$ has an infimum but no minimum, this tells us there is a limit point of $S$ that is not contained in $S$. All we can say is that there is a subsequence ${a_{n_k}}$ that converges to the infimum.
The more useful notion you may be looking for is the lim inf, which is the infimum of all possible subsequential limits. Note that in general
$$inf_{n} a_n le liminf_{n to infty} a_n$$
since the infimum is a subsequential limit. Whenever a sequence converges, it indeed converges to the lim inf.
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
answered Dec 3 '18 at 21:09
mysatellite
2,14221130
2,14221130
add a comment |
add a comment |
You can just create a series $a_n = left(frac{1}{n}right)$ defined on the real interval $(0,1]$. This way, the infimum is $0$, but there is no minimum because the interval is half-open.
answered Dec 3 '18 at 20:53
Thomas Lang
1624
add a comment |
You can just create a series $a_n = left(frac{1}{n}right)$ defined on the real interval $(0,1]$. This way, the infimum is $0$, but there is no minimum because the interval is half-open.
answered Dec 3 '18 at 20:53
Thomas Lang
1624
add a comment |
You can just create a series $a_n = left(frac{1}{n}right)$ defined on the real interval $(0,1]$. This way, the infimum is $0$, but there is no minimum because the interval is half-open.
answered Dec 3 '18 at 20:53
Thomas Lang
1624
You can just create a series $a_n = left(frac{1}{n}right)$ defined on the real interval $(0,1]$. This way, the infimum is $0$, but there is no minimum because the interval is half-open.
answered Dec 3 '18 at 20:53
Thomas Lang
1624
answered Dec 3 '18 at 20:53
Thomas Lang
1624
answered Dec 3 '18 at 20:53
Thomas Lang
1624
answered Dec 3 '18 at 20:53
Thomas Lang
1624
1624
add a comment |
add a comment |
3
The infimum is just the liminf, not necessarily limit.
– GNUSupporter 8964民主女神 地下教會
Dec 3 '18 at 20:38
4
The sequence ${a_n}$ defined by $$a_n = begin{cases} frac{1}{n} & text{$n$ is even,} \ 1 & text{otherwise}end{cases} $$ has infimum equal to zero, but has no limit.
– Xander Henderson
Dec 3 '18 at 20:44
2
@GNUSupporter8964民主女神地下教會 $$liminf a_n=inf a_n iff {a_n}text{ is monotonically decreasing }$$ Otherwise we always have $$ liminf a_nleinf a_n$$
– Daniele Tampieri
Dec 3 '18 at 20:55
1
@DanieleTampieri I left my previous comment too quickly. What about changing Xander Handerson's example from 1 to -1, in this way, liminf and inf are equal to -1, but $(a_n)$ is oscillating.
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 3:06
1
@GNUSupporter8964民主女神地下教會 I think it would be a nice counterexample. And also let me say that also I left my comment to quickly: a necessary premise for it would be "If ${a_n}$ is a converging sequence then" and the sign in the inequality between $liminf a_n$ and $inf a_n$ should be "$ge$", not "$le$".
– Daniele Tampieri
Dec 4 '18 at 5:59