MHB Compact Sets - Simple question about their nature .... ....

[Math Amateur]

Just a simple question regarding the nature of a compact set X in a metric space S:

Does X necessarily have to be infinite?

That is, are compact sets necessarily infinite?

Peter***EDIT***

Although I am most unsure about this it appears to me that a finite set can be compact since the set $$A \subset \mathbb{R}$$ where $$A = \{ 1,2,3 \}$$ is bounded and is also closed (since it contains all its limit points - it doesn't have any!) ...

I am most unsure of my example above, but cannot see the error in my analysis ... ...

Hope someone can clarify the above situation in the example ... ...

 

[Fallen Angel]

Hi Peter,

Compact sets can be finite or infinite, it doesn't mind.

But, when you are working in $\Bbb{R}^{n}$ with the usual topology every finite set is compact, so it's something more or less "trivial".

 

[Math Amateur]

Hi Peter,

Compact sets can be finite or infinite, it doesn't mind.

But, when you are working in $\Bbb{R}^{n}$ with the usual topology every finite set is compact, so it's something more or less "trivial".

Thanks so much for your help, Fallen Angel ... ...

Thanks also for pointing out that the statement that all finite sets are compact is compact only applies in certain spaces like $$\mathbb{R}^n$$ with the usual topology ...

Peter

 

[Fallen Angel]

Hi Peter,

I haven't said that =P

In fact, finite sets are compact in every topological space, maybe I should have explained this, sorry for the missunderstanding.

 

[caffeinemachine]

Thanks so much for your help, Fallen Angel ... ...

Thanks also for pointing out that the statement that all finite sets are compact is compact only applies in certain spaces like $$\mathbb{R}^n$$ with the usual topology ...

Peter

A subspace $S$ of a topological space $X$ is compact if and only if every open cover of $S$ admits a finite subcover.

From here it immediately follows that every finite subspace of a topological space is compact.

 

[Math Amateur]

A subspace $S$ of a topological space $X$ is compact if and only if every open cover of $S$ admits a finite subcover.

From here it immediately follows that every finite subspace of a topological space is compact.

Thanks caffeinemachine ... Appreciate your help ...

... But I am unsure why every open cover admitting a finite cover means the set is finite ...

Can you help?

Peter

 

[Euge]

Peter, you're thinking of the reverse implication: compact sets are finite. This is not true in general. Rather, you're trying to show that every finite subset $S$ of a toplogical space $X$ is compact. To do this, let $\mathcal{U}$ be an open covering of $S$. For sake of argument, let $S = \{s_1,\ldots, s_n\}$. Let $U_1,\ldots, U_n$ be elements of $\mathcal{U}$ such that $s_i \in U_i$ for $i = 1,2\ldots, n$. Then $\{U_1,\ldots, U_n\}$ is a finite subcover of $S$. Since $\mathcal{U}$ was arbitrary, $S$ is compact.

The following result is true: every compact discrete space is finite. To see this, let $X$ be a compact discrete space. Since $X$ is discrete, the singleton $\{x\}$ is open for all $x\in X$. Hence, the collection $\mathcal{U} = \{\{x\} : x\in X\}$ is an open covering of $X$. Since $X$ is compact, there exist points $x_1,\ldots, x_n \in X$ such that $X \subseteq \{x_1\} \cup \cdots \cup \{x_n\}$. Hence, $X = \{x_1,\ldots, x_n\}$.

 

[Math Amateur]

Peter, you're thinking of the reverse implication: compact sets are finite. This is not true in general. Rather, you're trying to show that every finite subset $S$ of a toplogical space $X$ is compact. To do this, let $\mathcal{U}$ be an open covering of $S$. For sake of argument, let $S = \{s_1,\ldots, s_n\}$. Let $U_1,\ldots, U_n$ be elements of $\mathcal{U}$ such that $s_i \in U_i$ for $i = 1,2\ldots, n$. Then $\{U_1,\ldots, U_n\}$ is a finite subcover of $S$. Since $\mathcal{U}$ was arbitrary, $S$ is compact.

The following result is true: every compact discrete space is finite. To see this, let $X$ be a compact discrete space. Since $X$ is discrete, the singleton $\{x\}$ is open for all $x\in X$. Hence, the collection $\mathcal{U} = \{\{x\} : x\in X\}$ is an open covering of $X$. Since $X$ is compact, there exist points $x_1,\ldots, x_n \in X$ such that $X \subseteq \{x_1\} \cup \cdots \cup \{x_n\}$. Hence, $X = \{x_1,\ldots, x_n\}$.

Thanks for for that clarification, Euge ...

Extremely helpful as usual ...

Peter