# Sequence compactness

In mathematics, a topological space is sequence- compact if every sequence has a convergent subsequence. Metric spaces are sequentially compact if and only if they are totally bounded and complete , i.e. compact . Hence subsets of the are sequentially compact (and compact) if and only if they are closed and bounded. There are topological spaces that are sequentially compact and not compact, and spaces that are not sequentially compact but compact. ${\ displaystyle \ mathbb {R} ^ {n}}$ ## Definitions

### Convergent sequences in topological spaces

If is a metric space, then a sequence converges with against if ${\ displaystyle (X, d)}$ ${\ displaystyle (x_ {i}) _ {i \ in \ mathbb {N}}}$ ${\ displaystyle x_ {i} \ in X}$ ${\ displaystyle x \ in X}$ ${\ displaystyle \ forall \ epsilon> 0 \ \ exists n \ in \ mathbb {N} \ \ forall i \ geq n: \ d (x_ {i}, x) <\ epsilon}$ .

This means that the sequence converges if and only if there is a natural number for every positive real number , so that all terms of the sequence starting from the -th term have a distance of that is smaller than . ${\ displaystyle x}$ ${\ displaystyle \ epsilon}$ ${\ displaystyle n}$ ${\ displaystyle n}$ ${\ displaystyle x}$ ${\ displaystyle \ epsilon}$ In any topological space, the surroundings take the place of the balls . Is a topological space, converges a series with against when it comes to any environment from one is so true for all . ${\ displaystyle B _ {\ epsilon} (x) = \ {y \ in X | d (y, x) <\ epsilon \}}$ ${\ displaystyle X}$ ${\ displaystyle (x_ {i}) _ {i \ in \ mathbb {N}}}$ ${\ displaystyle x_ {i} \ in X}$ ${\ displaystyle x \ in X}$ ${\ displaystyle U}$ ${\ displaystyle x}$ ${\ displaystyle n \ in \ mathbb {N}}$ ${\ displaystyle x_ {i} \ in U}$ ${\ displaystyle i \ geq n}$ ### Sequence compactness

A topological space is called sequence-compact if every sequence with contains a convergent subsequence . Correspondingly, a subspace is called sequence-compact if every sequence with has a convergent subsequence with limit in . ${\ displaystyle X}$ ${\ displaystyle (x_ {i}) _ {i \ in \ mathbb {N}}}$ ${\ displaystyle x_ {i} \ in X}$ ${\ displaystyle K \ subset X}$ ${\ displaystyle (x_ {i}) _ {i \ in \ mathbb {N}}}$ ${\ displaystyle x_ {i} \ in K}$ ${\ displaystyle K}$ ## Metric spaces

A metric space is sequentially compact if and only if it is compact. Because a metric space is compact precisely when it is totally limited and complete .

If a metric space is totally restricted, then every sequence contains a Cauchy sequence as a partial sequence. If it is also complete, this sequence converges. A compact metric space is consequently compact. More generally, any first-countable compact space is sequentially compact.

Conversely, if a metric space is sequence- compact, it must be totally restricted, since otherwise one and a sequence of points could be found, each with a distance of , and therefore would not have a convergent partial sequence. The space must also be complete, since a convergent subsequence of a Cauchy sequence must have the same limit as the original sequence. ${\ displaystyle \ epsilon> 0}$ ${\ displaystyle \ epsilon}$ ## Characteristics of compact spaces follow

A topological space is called countable-compact if every sequence has an accumulation point . Every sequentially compact space is countably compact. (The reverse does not apply.) In particular, every sequentially compact space is also weakly countably compact and pseudo-compact, since that is also every countably compact space. For metric spaces, compactness, sequence compactness and countable compactness always coincide.

## Examples

### A compact Hausdorff room that is not so compact

The set , provided with the discrete topology , is compact, therefore, according to Tychonoff's theorem , the set of all functions from the interval after , provided with the product topology , is also compact, and this space is Hausdorff-like . ${\ displaystyle \ {0.1 \}}$ ${\ displaystyle \ {0.1 \} ^ {[0.1]}}$ ${\ displaystyle [0,1]}$ ${\ displaystyle \ {0.1 \}}$ The fact that the product topology is provided means that a sequence of functions converges when it converges point by point. ${\ displaystyle \ {0.1 \} ^ {[0.1]}}$ However, this space is not compact:

A sequence of functions that does not contain a convergent subsequence can be defined as follows:

In the decimal analog notation in the binary system , the fractional part of a real number is an infinite sequence of zeros and ones.

The sequence is now defined as follows: is the -th decimal place of the number . ${\ displaystyle (a_ {n}) _ {n \ in \ mathbb {N}} \ (a_ {n} \ in \ {0,1 \} ^ {[0,1]})}$ ${\ displaystyle a_ {n} (x)}$ ${\ displaystyle n}$ ${\ displaystyle x}$ A number can now be defined for a partial sequence as follows . In the binary point representation, there is one in the -th position , if it is even and one in the other positions , if it is odd . This means that the sequence does not converge because the values ​​jump back and forth at the point . So the sequence cannot have a convergent subsequence. ${\ displaystyle (a_ {n_ {k}}) _ {n \ in \ mathbb {N}}}$ ${\ displaystyle y}$ ${\ displaystyle y}$ ${\ displaystyle n_ {k}}$ ${\ displaystyle 0}$ ${\ displaystyle k}$ ${\ displaystyle 1}$ ${\ displaystyle k}$ ${\ displaystyle 0}$ ${\ displaystyle (a_ {n_ {k}}) _ {n \ in \ mathbb {N}}}$ ${\ displaystyle y}$ ${\ displaystyle (a_ {n}) _ {n \ in \ mathbb {N}}}$ However, since the space is compact, the sequence has a convergent subnet . ${\ displaystyle (a_ {n})}$ ### A consequently compact space that is not compact

The first uncountable ordinal number (i.e. the uncountable set of all countable ordinal numbers ) is well ordered by the relation (the -relation) and therefore bears the topology of this order . ${\ displaystyle \ omega _ {1}}$ ${\ displaystyle [0, \ omega _ {1})}$ ${\ displaystyle <}$ ${\ displaystyle \ in}$ If now is a sequence of ordinal numbers, then the smallest ordinal number with the property that only finitely many sequence members are larger than it is an accumulation point of this sequence and the sequence can be thinned out to a convergent sequence. The space is therefore countably compact and consequently compact. ${\ displaystyle (a_ {n}) _ {n \ in \ mathbb {N}}}$ The family of open sets covers the set of all countable ordinals. A finite subfamily only contains countably many elements of . is therefore not compact. ${\ displaystyle \ bigcup _ {\ alpha <\ omega _ {1}} \ alpha = \ bigcup _ {\ alpha <\ omega _ {1}} \ {\ beta | \ beta <\ alpha \}}$ ${\ displaystyle \ omega _ {1}}$ ${\ displaystyle \ omega _ {1}}$ The fact that the set is not compact is because it does not contain the Limesordinal number . However, this is not the limit of a countable sequence, but only the limit of an uncountable network (e.g. given by all countable ordinal numbers in their natural order). ${\ displaystyle [0, \ omega _ {1})}$ ${\ displaystyle \ omega _ {1}}$ 