Regarding a metric space is compact, complete, connected

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Let $(X, d)$ be a metric space. Which is/are true?

1) if $X$ is countable, then $X$ is not connected.

2) If $X$ is countable, bounded and complete, then $X$ is compact.

3) if $X$ is compact, then $d$ is bounded.

My attempt :
Option ( 1) is false : if we consider singleton set is $X$.

For option 2) we know that if a space is complete and totally bounded then it is compact.
So I think 2 is true option

3) we know that the property of a metric space depends on the whole set and the metric and in a metric space a compact set is closed and bounded.
So I think option 3 is true.

Please help to solve this question. Thank you.

metric-spaces compactness connectedness complete-spaces

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edited Jul 14 at 18:21

Gonzalo Benavides

581317

asked Jul 14 at 17:27

Golam biswas

399

add a comment | 

up vote
1
down vote

favorite

Let $(X, d)$ be a metric space. Which is/are true?

1) if $X$ is countable, then $X$ is not connected.

2) If $X$ is countable, bounded and complete, then $X$ is compact.

3) if $X$ is compact, then $d$ is bounded.

My attempt :
Option ( 1) is false : if we consider singleton set is $X$.

For option 2) we know that if a space is complete and totally bounded then it is compact.
So I think 2 is true option

3) we know that the property of a metric space depends on the whole set and the metric and in a metric space a compact set is closed and bounded.
So I think option 3 is true.

Please help to solve this question. Thank you.

metric-spaces compactness connectedness complete-spaces

share|cite|improve this question

edited Jul 14 at 18:21

Gonzalo Benavides

581317

asked Jul 14 at 17:27

Golam biswas

399

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $(X, d)$ be a metric space. Which is/are true?

1) if $X$ is countable, then $X$ is not connected.

2) If $X$ is countable, bounded and complete, then $X$ is compact.

3) if $X$ is compact, then $d$ is bounded.

My attempt :
Option ( 1) is false : if we consider singleton set is $X$.

For option 2) we know that if a space is complete and totally bounded then it is compact.
So I think 2 is true option

3) we know that the property of a metric space depends on the whole set and the metric and in a metric space a compact set is closed and bounded.
So I think option 3 is true.

Please help to solve this question. Thank you.

metric-spaces compactness connectedness complete-spaces

share|cite|improve this question

edited Jul 14 at 18:21

Gonzalo Benavides

581317

asked Jul 14 at 17:27

Golam biswas

399

Let $(X, d)$ be a metric space. Which is/are true?

1) if $X$ is countable, then $X$ is not connected.

2) If $X$ is countable, bounded and complete, then $X$ is compact.

3) if $X$ is compact, then $d$ is bounded.

My attempt :
Option ( 1) is false : if we consider singleton set is $X$.

For option 2) we know that if a space is complete and totally bounded then it is compact.
So I think 2 is true option

3) we know that the property of a metric space depends on the whole set and the metric and in a metric space a compact set is closed and bounded.
So I think option 3 is true.

Please help to solve this question. Thank you.

metric-spaces compactness connectedness complete-spaces

share|cite|improve this question

edited Jul 14 at 18:21

Gonzalo Benavides

581317

asked Jul 14 at 17:27

Golam biswas

399

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 18:21

Gonzalo Benavides

581317

edited Jul 14 at 18:21

Gonzalo Benavides

581317

edited Jul 14 at 18:21

Gonzalo Benavides

581317

581317

asked Jul 14 at 17:27

Golam biswas

399

asked Jul 14 at 17:27

Golam biswas

399

asked Jul 14 at 17:27

Golam biswas

399

399

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

1. If finite sets are countable, then you're right. Otherwise, this statement is true.


2. This is false too. Take a countable set with the discrete metric.


3. Fix $x_0in X$. The map$$beginarraycccX&longrightarrow&mathbb R\x&mapsto& d(x,x_0)endarray$$is continuous and therefore (since $X$ is compact) it is bounded. If $R$ is an upper bound of its range, then $X$ is contained on the closed ball centered at $x_0$ with radius $R$ and therefore it is bounded.

share|cite|improve this answer

edited Jul 14 at 19:03

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  # Santos singleton set is not counble. Why?
  – Golam biswas
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If singleton is not countable then how can you say that option 1 is false?
  – Golam biswas
  Jul 14 at 18:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The most usual definition of countable set is: a set $C$ is countable if there is a bijection from $mathbb N$ to $C$. But, of course, if the definition of countable set that you wrotk with includes finite sets, then you're right. Concerning my statement that assertion 1. is false, did you notice that the word “false” in my answer is a link? I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir, I see your link. After seeing your link I assert that option 1 is right option. Am I right sir?
  – Golam biswas
  Jul 14 at 18:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Golambiswas No, you are wrong. There is no bijection between a finite set and $mathbb N$.
  – José Carlos Santos
  Jul 14 at 18:36
  

  
  
  
  

 | 
show 13 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

1. If finite sets are countable, then you're right. Otherwise, this statement is true.


2. This is false too. Take a countable set with the discrete metric.


3. Fix $x_0in X$. The map$$beginarraycccX&longrightarrow&mathbb R\x&mapsto& d(x,x_0)endarray$$is continuous and therefore (since $X$ is compact) it is bounded. If $R$ is an upper bound of its range, then $X$ is contained on the closed ball centered at $x_0$ with radius $R$ and therefore it is bounded.

share|cite|improve this answer

edited Jul 14 at 19:03

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  # Santos singleton set is not counble. Why?
  – Golam biswas
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If singleton is not countable then how can you say that option 1 is false?
  – Golam biswas
  Jul 14 at 18:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The most usual definition of countable set is: a set $C$ is countable if there is a bijection from $mathbb N$ to $C$. But, of course, if the definition of countable set that you wrotk with includes finite sets, then you're right. Concerning my statement that assertion 1. is false, did you notice that the word “false” in my answer is a link? I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir, I see your link. After seeing your link I assert that option 1 is right option. Am I right sir?
  – Golam biswas
  Jul 14 at 18:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Golambiswas No, you are wrong. There is no bijection between a finite set and $mathbb N$.
  – José Carlos Santos
  Jul 14 at 18:36
  

  
  
  
  

 | 
show 13 more comments

up vote
2
down vote



accepted

1. If finite sets are countable, then you're right. Otherwise, this statement is true.


2. This is false too. Take a countable set with the discrete metric.


3. Fix $x_0in X$. The map$$beginarraycccX&longrightarrow&mathbb R\x&mapsto& d(x,x_0)endarray$$is continuous and therefore (since $X$ is compact) it is bounded. If $R$ is an upper bound of its range, then $X$ is contained on the closed ball centered at $x_0$ with radius $R$ and therefore it is bounded.

share|cite|improve this answer

edited Jul 14 at 19:03

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  # Santos singleton set is not counble. Why?
  – Golam biswas
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If singleton is not countable then how can you say that option 1 is false?
  – Golam biswas
  Jul 14 at 18:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The most usual definition of countable set is: a set $C$ is countable if there is a bijection from $mathbb N$ to $C$. But, of course, if the definition of countable set that you wrotk with includes finite sets, then you're right. Concerning my statement that assertion 1. is false, did you notice that the word “false” in my answer is a link? I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir, I see your link. After seeing your link I assert that option 1 is right option. Am I right sir?
  – Golam biswas
  Jul 14 at 18:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Golambiswas No, you are wrong. There is no bijection between a finite set and $mathbb N$.
  – José Carlos Santos
  Jul 14 at 18:36
  

  
  
  
  

 | 
show 13 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

1. If finite sets are countable, then you're right. Otherwise, this statement is true.


2. This is false too. Take a countable set with the discrete metric.


3. Fix $x_0in X$. The map$$beginarraycccX&longrightarrow&mathbb R\x&mapsto& d(x,x_0)endarray$$is continuous and therefore (since $X$ is compact) it is bounded. If $R$ is an upper bound of its range, then $X$ is contained on the closed ball centered at $x_0$ with radius $R$ and therefore it is bounded.

share|cite|improve this answer

edited Jul 14 at 19:03

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

1. If finite sets are countable, then you're right. Otherwise, this statement is true.


2. This is false too. Take a countable set with the discrete metric.


3. Fix $x_0in X$. The map$$beginarraycccX&longrightarrow&mathbb R\x&mapsto& d(x,x_0)endarray$$is continuous and therefore (since $X$ is compact) it is bounded. If $R$ is an upper bound of its range, then $X$ is contained on the closed ball centered at $x_0$ with radius $R$ and therefore it is bounded.

share|cite|improve this answer

edited Jul 14 at 19:03

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 19:03

edited Jul 14 at 19:03

edited Jul 14 at 19:03

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

answered Jul 14 at 17:33

José Carlos Santos

114k1698177

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  # Santos singleton set is not counble. Why?
  – Golam biswas
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If singleton is not countable then how can you say that option 1 is false?
  – Golam biswas
  Jul 14 at 18:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The most usual definition of countable set is: a set $C$ is countable if there is a bijection from $mathbb N$ to $C$. But, of course, if the definition of countable set that you wrotk with includes finite sets, then you're right. Concerning my statement that assertion 1. is false, did you notice that the word “false” in my answer is a link? I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir, I see your link. After seeing your link I assert that option 1 is right option. Am I right sir?
  – Golam biswas
  Jul 14 at 18:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Golambiswas No, you are wrong. There is no bijection between a finite set and $mathbb N$.
  – José Carlos Santos
  Jul 14 at 18:36
  

  
  
  
  

 | 
show 13 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  # Santos singleton set is not counble. Why?
  – Golam biswas
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If singleton is not countable then how can you say that option 1 is false?
  – Golam biswas
  Jul 14 at 18:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The most usual definition of countable set is: a set $C$ is countable if there is a bijection from $mathbb N$ to $C$. But, of course, if the definition of countable set that you wrotk with includes finite sets, then you're right. Concerning my statement that assertion 1. is false, did you notice that the word “false” in my answer is a link? I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir, I see your link. After seeing your link I assert that option 1 is right option. Am I right sir?
  – Golam biswas
  Jul 14 at 18:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Golambiswas No, you are wrong. There is no bijection between a finite set and $mathbb N$.
  – José Carlos Santos
  Jul 14 at 18:36
  

  
  
  
  

# Santos singleton set is not counble. Why?
– Golam biswas
Jul 14 at 18:08

# Santos singleton set is not counble. Why?
– Golam biswas
Jul 14 at 18:08

If singleton is not countable then how can you say that option 1 is false?
– Golam biswas
Jul 14 at 18:09

If singleton is not countable then how can you say that option 1 is false?
– Golam biswas
Jul 14 at 18:09

2

2

The most usual definition of countable set is: a set $C$ is countable if there is a bijection from $mathbb N$ to $C$. But, of course, if the definition of countable set that you wrotk with includes finite sets, then you're right. Concerning my statement that assertion 1. is false, did you notice that the word “false” in my answer is a link? I've edited my answer.
– José Carlos Santos
Jul 14 at 18:12

The most usual definition of countable set is: a set $C$ is countable if there is a bijection from $mathbb N$ to $C$. But, of course, if the definition of countable set that you wrotk with includes finite sets, then you're right. Concerning my statement that assertion 1. is false, did you notice that the word “false” in my answer is a link? I've edited my answer.
– José Carlos Santos
Jul 14 at 18:12

Yes sir, I see your link. After seeing your link I assert that option 1 is right option. Am I right sir?
– Golam biswas
Jul 14 at 18:20

Yes sir, I see your link. After seeing your link I assert that option 1 is right option. Am I right sir?
– Golam biswas
Jul 14 at 18:20

1

1

@Golambiswas No, you are wrong. There is no bijection between a finite set and $mathbb N$.
– José Carlos Santos
Jul 14 at 18:36

@Golambiswas No, you are wrong. There is no bijection between a finite set and $mathbb N$.
– José Carlos Santos
Jul 14 at 18:36

 | 
show 13 more comments

 

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