Regarding a metric space is compact, complete, connected

Question

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.

José Carlos Santos · Accepted Answer · 2018-07-14T19:03:22.307

2

If finite sets are countable, then you're right. Otherwise, this statement is true.
This is false too. Take a countable set with the discrete metric.
Fix $x_0\in X$. The map$$\begin{array}{ccc}X&\longrightarrow&\mathbb R\\x&\mapsto& d(x,x_0)\end{array}$$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.

edited Jul 14 '18 at 19:03

answered Jul 14 '18 at 17:33

José Carlos Santos

440,053

Santos singleton set is not counble. Why? – Golam biswas Jul 14 '18 at 18:08
If singleton is not countable then how can you say that option 1 is false? – Golam biswas Jul 14 '18 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 '18 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 '18 at 18:20
@Golambiswas As I wrote in my previous comment, this depends upon the defintion of countable that you use. So, please tell me how is it that you define “countable set”. – José Carlos Santos Jul 14 '18 at 18:21
Yes sir, I use same definition of countable set. – Golam biswas Jul 14 '18 at 18:24
@Golambiswas The same as mine? – José Carlos Santos Jul 14 '18 at 18:24
Yes sir. I agree on your definition. I use this definition as you mentioned. – Golam biswas Jul 14 '18 at 18:27
@Golambiswas Then a singleton is not countable. – José Carlos Santos Jul 14 '18 at 18:27
Then tell me your opinion about the option 1 is false or correct? Plz – Golam biswas Jul 14 '18 at 18:30
@Golambiswas Given the definition of countable that we both use, the first option is false. That's what I wrote in my answer. – José Carlos Santos Jul 14 '18 at 18:31
But sir , a singleton set is a finite set. And a finite set is countable. Am I write sir? – Golam biswas Jul 14 '18 at 18:33
1

@Golambiswas No, you are wrong. There is no bijection between a finite set and $\mathbb N$. – José Carlos Santos Jul 14 '18 at 18:36
Okkk sir. Thank you so much. – Golam biswas Jul 14 '18 at 18:38
@ Santos sir, can you give an example regarding option (1)? – Golam biswas Jul 14 '18 at 19:01
@Golambiswas No, I cannot. I got confused and I wrote that the statement is false, but it is true (the link is correct, though). I've edited my answer. Sorry about that. – José Carlos Santos Jul 14 '18 at 19:04
I told this when you given the link.. Okk sir thanku so much. – Golam biswas Jul 14 '18 at 19:07
@Golambiswas Now I see that you did that. Again, I'm sorry. – José Carlos Santos Jul 14 '18 at 19:08

Regarding a metric space is compact, complete, connected

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