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A set that is bounded below possess a subsequence that converges to its infimum.

Subsequence that converges to $lim text{inf}$compact set always contains its supremum and infimumShow that a sequence is bounded if and only if there exists a K $inmathbb{R}$ such that $| a_n | leq K$ $forall nin mathbb{N}$.IF a set is bounded below, then its complement is not bounded belowImplicit Fact about Bounded Monotone Sequence convergesProving $inf F = - sup E$Definition of a sequence not bounded below.Show that every bounded sequence in $mathbb{R}^n$ has a convergent subsequence.Infimum of a sequence bounded from belowIs “bounded” required for the definition below

0

$begingroup$

Assume $S subset mathbb{R}$ is nonempty and bounded below. Show that there exists a sequence $(a_n)_{ nin mathbb{N}}in S^{mathbb{N}}$ of elements of $S$ converging to $inf(S)$.

real-analysis general-topology

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edited yesterday

Math Enthusiast

asked yesterday

Math Enthusiast Math Enthusiast

103

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have any characterizations for the infimum or just the definition, that is, the largest lower bound?
  $endgroup$
  – clark
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think that the idea of compactness is useful here? The question doesn't mention compactness of $S$ anywhere at all as of now.
  $endgroup$
  – Error 404
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Error404 I thought it might be useful.
  $endgroup$
  – Math Enthusiast
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @clark just the definition.
  $endgroup$
  – Math Enthusiast
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

Assume $S subset mathbb{R}$ is nonempty and bounded below. Show that there exists a sequence $(a_n)_{ nin mathbb{N}}in S^{mathbb{N}}$ of elements of $S$ converging to $inf(S)$.

real-analysis general-topology

share|cite|improve this question

edited yesterday

Math Enthusiast

asked yesterday

Math Enthusiast Math Enthusiast

103

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have any characterizations for the infimum or just the definition, that is, the largest lower bound?
  $endgroup$
  – clark
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think that the idea of compactness is useful here? The question doesn't mention compactness of $S$ anywhere at all as of now.
  $endgroup$
  – Error 404
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Error404 I thought it might be useful.
  $endgroup$
  – Math Enthusiast
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @clark just the definition.
  $endgroup$
  – Math Enthusiast
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Assume $S subset mathbb{R}$ is nonempty and bounded below. Show that there exists a sequence $(a_n)_{ nin mathbb{N}}in S^{mathbb{N}}$ of elements of $S$ converging to $inf(S)$.

real-analysis general-topology

share|cite|improve this question

edited yesterday

Math Enthusiast

asked yesterday

Math Enthusiast Math Enthusiast

103

$endgroup$

share|cite|improve this question

edited yesterday

Math Enthusiast

asked yesterday

Math Enthusiast Math Enthusiast

103

share|cite|improve this question

edited yesterday

Math Enthusiast

asked yesterday

Math Enthusiast Math Enthusiast

103

asked yesterday

Math Enthusiast Math Enthusiast

103

asked yesterday

Math Enthusiast Math Enthusiast

103

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $s$ denote the infimum of $S$. For each $epsilon > 0$, the quantity $s+epsilon$ cannot be a lower-bound for $S$. Hence, there exists (see the note below if not clear) $x_epsilon in S$ such that
$$
sleq x_epsilon < s+epsilon.
$$
Given $n in mathbb{N}$ and using $epsilon = 1/n$ in the above, we obtain $x_n in S$ such that
$$
s leq x_n < s + frac{1}{n}.
$$
By the squeeze theorem, it follows that $x_n to s$ as $n to infty$.

Note: If $s+epsilon$ is not a lower bound for $S$, this means that the statement
$$
s+epsilon leq x, quad forall x in S
$$
is false. Put otherwise, there must exist $x in S$ for which the above fails, i.e. such that $x < s+ epsilon$. But, because $s$ is a lower bound for $S ni x$, we still have $s leq x$.

share|cite|improve this answer

edited yesterday

answered yesterday

rolandcyprolandcyp

46739

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. What do you mean by $epsilon:=n?$
  $endgroup$
  – Math Enthusiast
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The procedure used to construct $x_epsilon$ works for each $epsilon > 0$. Hence, it works for every natural number $n in mathbb{N}$. Repeating this for each $n$ gives you the sequence $(x_n)$. Hope this helps.
  $endgroup$
  – rolandcyp
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rolandcyp I believe MathEnthusiant's comment is about your definition of $epsilon$. Should it not be $epsilon = frac{1}{n}$ instead of $epsilon = n$?
  $endgroup$
  – John Omielan
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you're totally right. Thanks!
  $endgroup$
  – rolandcyp
  yesterday
  

  
  
  
  

add a comment | 

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1

$begingroup$

Let $s$ denote the infimum of $S$. For each $epsilon > 0$, the quantity $s+epsilon$ cannot be a lower-bound for $S$. Hence, there exists (see the note below if not clear) $x_epsilon in S$ such that
$$
sleq x_epsilon < s+epsilon.
$$
Given $n in mathbb{N}$ and using $epsilon = 1/n$ in the above, we obtain $x_n in S$ such that
$$
s leq x_n < s + frac{1}{n}.
$$
By the squeeze theorem, it follows that $x_n to s$ as $n to infty$.

Note: If $s+epsilon$ is not a lower bound for $S$, this means that the statement
$$
s+epsilon leq x, quad forall x in S
$$
is false. Put otherwise, there must exist $x in S$ for which the above fails, i.e. such that $x < s+ epsilon$. But, because $s$ is a lower bound for $S ni x$, we still have $s leq x$.

share|cite|improve this answer

edited yesterday

answered yesterday

rolandcyprolandcyp

46739

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. What do you mean by $epsilon:=n?$
  $endgroup$
  – Math Enthusiast
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The procedure used to construct $x_epsilon$ works for each $epsilon > 0$. Hence, it works for every natural number $n in mathbb{N}$. Repeating this for each $n$ gives you the sequence $(x_n)$. Hope this helps.
  $endgroup$
  – rolandcyp
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rolandcyp I believe MathEnthusiant's comment is about your definition of $epsilon$. Should it not be $epsilon = frac{1}{n}$ instead of $epsilon = n$?
  $endgroup$
  – John Omielan
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you're totally right. Thanks!
  $endgroup$
  – rolandcyp
  yesterday
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $s$ denote the infimum of $S$. For each $epsilon > 0$, the quantity $s+epsilon$ cannot be a lower-bound for $S$. Hence, there exists (see the note below if not clear) $x_epsilon in S$ such that
$$
sleq x_epsilon < s+epsilon.
$$
Given $n in mathbb{N}$ and using $epsilon = 1/n$ in the above, we obtain $x_n in S$ such that
$$
s leq x_n < s + frac{1}{n}.
$$
By the squeeze theorem, it follows that $x_n to s$ as $n to infty$.

Note: If $s+epsilon$ is not a lower bound for $S$, this means that the statement
$$
s+epsilon leq x, quad forall x in S
$$
is false. Put otherwise, there must exist $x in S$ for which the above fails, i.e. such that $x < s+ epsilon$. But, because $s$ is a lower bound for $S ni x$, we still have $s leq x$.

share|cite|improve this answer

edited yesterday

answered yesterday

rolandcyprolandcyp

46739

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. What do you mean by $epsilon:=n?$
  $endgroup$
  – Math Enthusiast
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The procedure used to construct $x_epsilon$ works for each $epsilon > 0$. Hence, it works for every natural number $n in mathbb{N}$. Repeating this for each $n$ gives you the sequence $(x_n)$. Hope this helps.
  $endgroup$
  – rolandcyp
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rolandcyp I believe MathEnthusiant's comment is about your definition of $epsilon$. Should it not be $epsilon = frac{1}{n}$ instead of $epsilon = n$?
  $endgroup$
  – John Omielan
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you're totally right. Thanks!
  $endgroup$
  – rolandcyp
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Let $s$ denote the infimum of $S$. For each $epsilon > 0$, the quantity $s+epsilon$ cannot be a lower-bound for $S$. Hence, there exists (see the note below if not clear) $x_epsilon in S$ such that
$$
sleq x_epsilon < s+epsilon.
$$
Given $n in mathbb{N}$ and using $epsilon = 1/n$ in the above, we obtain $x_n in S$ such that
$$
s leq x_n < s + frac{1}{n}.
$$
By the squeeze theorem, it follows that $x_n to s$ as $n to infty$.

Note: If $s+epsilon$ is not a lower bound for $S$, this means that the statement
$$
s+epsilon leq x, quad forall x in S
$$
is false. Put otherwise, there must exist $x in S$ for which the above fails, i.e. such that $x < s+ epsilon$. But, because $s$ is a lower bound for $S ni x$, we still have $s leq x$.

share|cite|improve this answer

edited yesterday

answered yesterday

rolandcyprolandcyp

46739

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

rolandcyprolandcyp

46739

answered yesterday

rolandcyprolandcyp

46739

answered yesterday

rolandcyprolandcyp

46739