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Prove that a subset C of $mathbb R^n$ is closed if and only if it contains all its limit points.

3

$begingroup$

Prove that a subset C of $mathbb R^n$ is closed if and only if it contains all its limit points.

A closed set is defined by a set of all boundary points

My professor said
"We may prove that C is not closed if and only if $C^c$ has a limit point of C.
To prove this, it suffices to show that a point x in $C^c$ is a boundary point of C if and only it is a limit point of C." and he left it as an exercise.

I am a beginner of analysis and I want you to check if my proof is okay.

This is how I prove:

If part)
Let $xin C^c$ and x is a limit point of C. Then there is a sequence ${x_k}$ of distinct points in C such that $0<Vert x_k -x Vert<1/k$. Thus $N'(x;r)bigcap C$ is infinite. It follows that $N(x;r)bigcap Cneqemptyset$. Since $xin C^c$, we get $N(x;r)bigcap C^cneqemptyset$. Hence x is a boundary point of C.

Only if part)
Let $xin C^c$. Since x is a boundary point of C, there exists $x_kin C$ such that $x=lim_{kto infty} x_k$. Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}.$ Then $Vert x- x_{k_j} Vert > Vert x- x_{k_{j+1}} Vert$, so ${x_{k_j}}$ converges to x. Thus x is a limit point.

Please tell me if there is wrong or insufficient.
Is it okay just to say that "Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}$"?

analysis

share|cite|improve this question

edited May 5 '16 at 0:54

Chloe

asked May 4 '16 at 17:05

ChloeChloe

956

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you give an example of ${x_{k_j}}$ where $x_{k_j} ne x_{k_j}$?
  $endgroup$
  – T. Bongers
  May 4 '16 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}$, not j but l.
  $endgroup$
  – Chloe
  May 4 '16 at 17:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How is a closed set defined?
  $endgroup$
  – velut luna
  May 4 '16 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A closed set is defined by a set of all boundary points!
  $endgroup$
  – Chloe
  May 4 '16 at 23:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You mean a closed set is defined to be a set that contains all its boundary points?
  $endgroup$
  – velut luna
  May 5 '16 at 1:10
  

  
  
  
  

 | 
show 1 more comment

3

$begingroup$

Prove that a subset C of $mathbb R^n$ is closed if and only if it contains all its limit points.

A closed set is defined by a set of all boundary points

My professor said
"We may prove that C is not closed if and only if $C^c$ has a limit point of C.
To prove this, it suffices to show that a point x in $C^c$ is a boundary point of C if and only it is a limit point of C." and he left it as an exercise.

I am a beginner of analysis and I want you to check if my proof is okay.

This is how I prove:

If part)
Let $xin C^c$ and x is a limit point of C. Then there is a sequence ${x_k}$ of distinct points in C such that $0<Vert x_k -x Vert<1/k$. Thus $N'(x;r)bigcap C$ is infinite. It follows that $N(x;r)bigcap Cneqemptyset$. Since $xin C^c$, we get $N(x;r)bigcap C^cneqemptyset$. Hence x is a boundary point of C.

Only if part)
Let $xin C^c$. Since x is a boundary point of C, there exists $x_kin C$ such that $x=lim_{kto infty} x_k$. Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}.$ Then $Vert x- x_{k_j} Vert > Vert x- x_{k_{j+1}} Vert$, so ${x_{k_j}}$ converges to x. Thus x is a limit point.

Please tell me if there is wrong or insufficient.
Is it okay just to say that "Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}$"?

analysis

share|cite|improve this question

edited May 5 '16 at 0:54

Chloe

asked May 4 '16 at 17:05

ChloeChloe

956

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you give an example of ${x_{k_j}}$ where $x_{k_j} ne x_{k_j}$?
  $endgroup$
  – T. Bongers
  May 4 '16 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}$, not j but l.
  $endgroup$
  – Chloe
  May 4 '16 at 17:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How is a closed set defined?
  $endgroup$
  – velut luna
  May 4 '16 at 18:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A closed set is defined by a set of all boundary points!
  $endgroup$
  – Chloe
  May 4 '16 at 23:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You mean a closed set is defined to be a set that contains all its boundary points?
  $endgroup$
  – velut luna
  May 5 '16 at 1:10
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Prove that a subset C of $mathbb R^n$ is closed if and only if it contains all its limit points.

A closed set is defined by a set of all boundary points

My professor said
"We may prove that C is not closed if and only if $C^c$ has a limit point of C.
To prove this, it suffices to show that a point x in $C^c$ is a boundary point of C if and only it is a limit point of C." and he left it as an exercise.

I am a beginner of analysis and I want you to check if my proof is okay.

This is how I prove:

If part)
Let $xin C^c$ and x is a limit point of C. Then there is a sequence ${x_k}$ of distinct points in C such that $0<Vert x_k -x Vert<1/k$. Thus $N'(x;r)bigcap C$ is infinite. It follows that $N(x;r)bigcap Cneqemptyset$. Since $xin C^c$, we get $N(x;r)bigcap C^cneqemptyset$. Hence x is a boundary point of C.

Only if part)
Let $xin C^c$. Since x is a boundary point of C, there exists $x_kin C$ such that $x=lim_{kto infty} x_k$. Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}.$ Then $Vert x- x_{k_j} Vert > Vert x- x_{k_{j+1}} Vert$, so ${x_{k_j}}$ converges to x. Thus x is a limit point.

Please tell me if there is wrong or insufficient.
Is it okay just to say that "Choose ${x_{k_j}}$ where $x_{k_j}neq x_{k_l}$"?

analysis

share|cite|improve this question

edited May 5 '16 at 0:54

Chloe

asked May 4 '16 at 17:05

ChloeChloe

956

$endgroup$

share|cite|improve this question

edited May 5 '16 at 0:54

Chloe

asked May 4 '16 at 17:05

ChloeChloe

956

share|cite|improve this question

edited May 5 '16 at 0:54

Chloe

asked May 4 '16 at 17:05

ChloeChloe

956

asked May 4 '16 at 17:05

ChloeChloe

956

asked May 4 '16 at 17:05

ChloeChloe

956

1 Answer
1

active

oldest

votes

1

$begingroup$

For topological space, the usual definition of a closed set is that it contains all its limit points. So your statement to prove is in fact a definition.

OK so here, instead, we define a set to be closed if it contains all its boundary points.

I think your only if part can be simplified.

If $x$ is a boundary point of $C$ and $xin C^c$, then by definition there exists a sequence ${x_k}$ where $x_k ne x$ for all $k$ and
$$lim_{krightarrowinfty}x_k=x$$.

In fact, I think the proof is much simpler if you use, instead of sequence, the following definition of limit point: $x$ is a limit point of $C$ if every neighbourhood of $x$ contains at least one point of $C$ different from $x$ itself.

Then the proof is much simpler:

Let $V_x$ be any neighbourhood of $x$.
$$C text{ is not closed} iff exists x in partial C text{ and } x in C^c$$
$$iff exists x'ne x, x' in C, x'in V_x$$
$$iff x text{ is a limit point of }C$$

share|cite|improve this answer

edited May 5 '16 at 12:51

answered May 5 '16 at 10:33

velut lunavelut luna

8,08111240

$endgroup$

add a comment | 

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1

$begingroup$

For topological space, the usual definition of a closed set is that it contains all its limit points. So your statement to prove is in fact a definition.

OK so here, instead, we define a set to be closed if it contains all its boundary points.

I think your only if part can be simplified.

If $x$ is a boundary point of $C$ and $xin C^c$, then by definition there exists a sequence ${x_k}$ where $x_k ne x$ for all $k$ and
$$lim_{krightarrowinfty}x_k=x$$.

In fact, I think the proof is much simpler if you use, instead of sequence, the following definition of limit point: $x$ is a limit point of $C$ if every neighbourhood of $x$ contains at least one point of $C$ different from $x$ itself.

Then the proof is much simpler:

Let $V_x$ be any neighbourhood of $x$.
$$C text{ is not closed} iff exists x in partial C text{ and } x in C^c$$
$$iff exists x'ne x, x' in C, x'in V_x$$
$$iff x text{ is a limit point of }C$$

share|cite|improve this answer

edited May 5 '16 at 12:51

answered May 5 '16 at 10:33

velut lunavelut luna

8,08111240

$endgroup$

add a comment | 

1

$begingroup$

For topological space, the usual definition of a closed set is that it contains all its limit points. So your statement to prove is in fact a definition.

OK so here, instead, we define a set to be closed if it contains all its boundary points.

I think your only if part can be simplified.

If $x$ is a boundary point of $C$ and $xin C^c$, then by definition there exists a sequence ${x_k}$ where $x_k ne x$ for all $k$ and
$$lim_{krightarrowinfty}x_k=x$$.

In fact, I think the proof is much simpler if you use, instead of sequence, the following definition of limit point: $x$ is a limit point of $C$ if every neighbourhood of $x$ contains at least one point of $C$ different from $x$ itself.

Then the proof is much simpler:

Let $V_x$ be any neighbourhood of $x$.
$$C text{ is not closed} iff exists x in partial C text{ and } x in C^c$$
$$iff exists x'ne x, x' in C, x'in V_x$$
$$iff x text{ is a limit point of }C$$

share|cite|improve this answer

edited May 5 '16 at 12:51

answered May 5 '16 at 10:33

velut lunavelut luna

8,08111240

$endgroup$

add a comment | 

$begingroup$

For topological space, the usual definition of a closed set is that it contains all its limit points. So your statement to prove is in fact a definition.

OK so here, instead, we define a set to be closed if it contains all its boundary points.

I think your only if part can be simplified.

If $x$ is a boundary point of $C$ and $xin C^c$, then by definition there exists a sequence ${x_k}$ where $x_k ne x$ for all $k$ and
$$lim_{krightarrowinfty}x_k=x$$.

In fact, I think the proof is much simpler if you use, instead of sequence, the following definition of limit point: $x$ is a limit point of $C$ if every neighbourhood of $x$ contains at least one point of $C$ different from $x$ itself.

Then the proof is much simpler:

Let $V_x$ be any neighbourhood of $x$.
$$C text{ is not closed} iff exists x in partial C text{ and } x in C^c$$
$$iff exists x'ne x, x' in C, x'in V_x$$
$$iff x text{ is a limit point of }C$$

share|cite|improve this answer

edited May 5 '16 at 12:51

answered May 5 '16 at 10:33

velut lunavelut luna

8,08111240

$endgroup$

share|cite|improve this answer

edited May 5 '16 at 12:51

answered May 5 '16 at 10:33

velut lunavelut luna

8,08111240

answered May 5 '16 at 10:33

velut lunavelut luna

8,08111240

answered May 5 '16 at 10:33

velut lunavelut luna

8,08111240