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How to prove that the convergent value of a sequence is unique?

1

$begingroup$

I need a little support in the following proposition:

Let $(x_n)$ be a convergene sequence in $mathbb{R}$ and $x_n rightarrow a $ and $x_n rightarrow b $ , then $a=b$

My proof is:

Consider $|a-b| = |a-x_n + x_n-b| le |a-x_n| + |x_n-b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$
so it follows tat $|a-b| <epsilon$ then $a = b$

is that correct?

real-analysis

share|cite|improve this question

edited Mar 26 at 0:46

José Marín

asked Mar 26 at 0:40

José MarínJosé Marín

250211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The idea is correct but the proof is incomplete.
  $endgroup$
  – mfl
  Mar 26 at 0:44
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is missing?
  $endgroup$
  – José Marín
  Mar 26 at 0:44
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You haven't ever defined a symbol. You didn't make any assumptions about $epsilon$, you haven't connected it with the sequence ${x_n}$, you haven't stated that $|a - b| < epsilon$ holds for all positive $epsilon$... Essentially all of the explanation of the proof is missing here.
  $endgroup$
  – T. Bongers
  Mar 26 at 0:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. But if I add those details it will be ok, right?
  $endgroup$
  – José Marín
  Mar 26 at 0:53
  

  
  
  
  

add a comment | 

1

$begingroup$

I need a little support in the following proposition:

Let $(x_n)$ be a convergene sequence in $mathbb{R}$ and $x_n rightarrow a $ and $x_n rightarrow b $ , then $a=b$

My proof is:

Consider $|a-b| = |a-x_n + x_n-b| le |a-x_n| + |x_n-b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$
so it follows tat $|a-b| <epsilon$ then $a = b$

is that correct?

real-analysis

share|cite|improve this question

edited Mar 26 at 0:46

José Marín

asked Mar 26 at 0:40

José MarínJosé Marín

250211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The idea is correct but the proof is incomplete.
  $endgroup$
  – mfl
  Mar 26 at 0:44
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is missing?
  $endgroup$
  – José Marín
  Mar 26 at 0:44
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You haven't ever defined a symbol. You didn't make any assumptions about $epsilon$, you haven't connected it with the sequence ${x_n}$, you haven't stated that $|a - b| < epsilon$ holds for all positive $epsilon$... Essentially all of the explanation of the proof is missing here.
  $endgroup$
  – T. Bongers
  Mar 26 at 0:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. But if I add those details it will be ok, right?
  $endgroup$
  – José Marín
  Mar 26 at 0:53
  

  
  
  
  

add a comment | 

$begingroup$

I need a little support in the following proposition:

Let $(x_n)$ be a convergene sequence in $mathbb{R}$ and $x_n rightarrow a $ and $x_n rightarrow b $ , then $a=b$

My proof is:

Consider $|a-b| = |a-x_n + x_n-b| le |a-x_n| + |x_n-b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$
so it follows tat $|a-b| <epsilon$ then $a = b$

is that correct?

real-analysis

share|cite|improve this question

edited Mar 26 at 0:46

José Marín

asked Mar 26 at 0:40

José MarínJosé Marín

250211

$endgroup$

share|cite|improve this question

edited Mar 26 at 0:46

José Marín

asked Mar 26 at 0:40

José MarínJosé Marín

250211

share|cite|improve this question

edited Mar 26 at 0:46

José Marín

asked Mar 26 at 0:40

José MarínJosé Marín

250211

asked Mar 26 at 0:40

José MarínJosé Marín

250211

asked Mar 26 at 0:40

José MarínJosé Marín

250211

1 Answer
1

active

oldest

votes

2

$begingroup$

The idea behind your proof is correct but it lacks a certain formality. Namely, you have not specified the value of $n$ you are using when discussing $x_n$! Below, we make the proof formal.

Let $epsilon > 0$ be given, our goal is to show that $|a-b| < epsilon$. Since $x_n to a$, there exists $N_1 geq 1$ such that
$$
|x_n- a| < frac{epsilon}{2}
$$
for all $n geq N_1$. Similarly, because $x_n to b$, there is $N_2 in mathbb{N}$ with the property that
$$|x_n - b| <frac{epsilon}{2}
$$
for all $n geq N_2$. Now, if $N := max(N_1,N_2)$ and $n geq N$, we have by the triangle inequality that
begin{align*}
|a-b| leq |a-x_n| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.
end{align*}
Hence, $|a-b| < epsilon$. Using that $epsilon >0$ was arbitrary, it follows that $a=b$.

share|cite|improve this answer

answered Mar 26 at 1:16

rolandcyprolandcyp

2,149422

$endgroup$

add a comment | 

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2

$begingroup$

The idea behind your proof is correct but it lacks a certain formality. Namely, you have not specified the value of $n$ you are using when discussing $x_n$! Below, we make the proof formal.

Let $epsilon > 0$ be given, our goal is to show that $|a-b| < epsilon$. Since $x_n to a$, there exists $N_1 geq 1$ such that
$$
|x_n- a| < frac{epsilon}{2}
$$
for all $n geq N_1$. Similarly, because $x_n to b$, there is $N_2 in mathbb{N}$ with the property that
$$|x_n - b| <frac{epsilon}{2}
$$
for all $n geq N_2$. Now, if $N := max(N_1,N_2)$ and $n geq N$, we have by the triangle inequality that
begin{align*}
|a-b| leq |a-x_n| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.
end{align*}
Hence, $|a-b| < epsilon$. Using that $epsilon >0$ was arbitrary, it follows that $a=b$.

share|cite|improve this answer

answered Mar 26 at 1:16

rolandcyprolandcyp

2,149422

$endgroup$

add a comment | 

2

$begingroup$

The idea behind your proof is correct but it lacks a certain formality. Namely, you have not specified the value of $n$ you are using when discussing $x_n$! Below, we make the proof formal.

Let $epsilon > 0$ be given, our goal is to show that $|a-b| < epsilon$. Since $x_n to a$, there exists $N_1 geq 1$ such that
$$
|x_n- a| < frac{epsilon}{2}
$$
for all $n geq N_1$. Similarly, because $x_n to b$, there is $N_2 in mathbb{N}$ with the property that
$$|x_n - b| <frac{epsilon}{2}
$$
for all $n geq N_2$. Now, if $N := max(N_1,N_2)$ and $n geq N$, we have by the triangle inequality that
begin{align*}
|a-b| leq |a-x_n| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.
end{align*}
Hence, $|a-b| < epsilon$. Using that $epsilon >0$ was arbitrary, it follows that $a=b$.

share|cite|improve this answer

answered Mar 26 at 1:16

rolandcyprolandcyp

2,149422

$endgroup$

add a comment | 

$begingroup$

The idea behind your proof is correct but it lacks a certain formality. Namely, you have not specified the value of $n$ you are using when discussing $x_n$! Below, we make the proof formal.

Let $epsilon > 0$ be given, our goal is to show that $|a-b| < epsilon$. Since $x_n to a$, there exists $N_1 geq 1$ such that
$$
|x_n- a| < frac{epsilon}{2}
$$
for all $n geq N_1$. Similarly, because $x_n to b$, there is $N_2 in mathbb{N}$ with the property that
$$|x_n - b| <frac{epsilon}{2}
$$
for all $n geq N_2$. Now, if $N := max(N_1,N_2)$ and $n geq N$, we have by the triangle inequality that
begin{align*}
|a-b| leq |a-x_n| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.
end{align*}
Hence, $|a-b| < epsilon$. Using that $epsilon >0$ was arbitrary, it follows that $a=b$.

share|cite|improve this answer

answered Mar 26 at 1:16

rolandcyprolandcyp

2,149422

$endgroup$

share|cite|improve this answer

answered Mar 26 at 1:16

rolandcyprolandcyp

2,149422

answered Mar 26 at 1:16

rolandcyprolandcyp

2,149422

answered Mar 26 at 1:16

rolandcyprolandcyp

2,149422