The definition I have for sequential compactness goes like this:

A subspace $ K \subseteq X $, $X$ a metric space, is sequentially compact if all sequences $(x_n)$, $x_n \in K$ have a convergent subsequence $(x_{n_i})$, $x_{n_i} \rightarrow x \in K$.

My interpretation of this is the following statement:

Let $ K \subseteq X $, $X$ a metric space. Then:

all sequences $(x_n)$, $x_n \in K$ have a convergent subsequence $(x_{n_i})$, $x_{n_i} \rightarrow x \in K$ $\implies$ $K$ is compact.

But I was wondering if/when this is true in the other direction? I thought that if $K$ is compact and it contains some sequence, then that sequence must have a convergent subsequence? But perhaps there is no guarantee that $K$ contains a sequence? But then if $K$ is non-empty, it must have points in it which form some sort of sequence?

I'd be grateful if someone could clarify this for me. Many thanks.


Definitions in math are biconditional statements. Confusingly, they are often not stated that way. However, it is the case that

$$\text{All sequences $(x_n)$ with elements in $K$ have a convergent subsequence $(x_{n_i})$ $\Leftrightarrow$ $K$ is compact}$$

The case of the empty set is interesting. As an analogy, suppose I said "all the irrational integers are zero". That statement is nonsense, clearly, but it's also true. Since there are no irrational integers, it doesn't matter what statement I make about them. Such a statement is said to be "vacuously true".

So if $K$ is empty, it is true that every sequence in $K$ contains a convergent subsequence, but it's vacuously true since $K$ does not contain any sequences.

  • $\begingroup$ Some professors always use "iff" in definitions, but I always find it to be pedantic XD $\endgroup$ – Violapterin Oct 18 '19 at 15:52

Suppose $X$ is a metric space and that $k \subset X$. A set $K$ of a topological space is said to be compact, when for every collection of open sets $(U_i)_{i \in I}$ such that $K \subset \cup_{i \in I} U_i$ there exists a finite number of indexes, said $i_1, \cdots, i_r$ such that $K \subset U_{i_1} \cup \cdots \cup U_{i_r}$. This is called the Borel-Lebesgue property. Now, it can be shown that, inside metric spaces, this property is equivalent to the property of sequential compactness. Therefore $K$ is compact (almost by definition) if and only if $K$ is sequentially compact.


In general compactness and sequential compactness are not equivalent, neither one necessarily implies the other. This equivalence is true, however, for metric spaces.

So a metric space is compact if and only if it is sequentially compact.

For this reason, it is useful and helpful not to think about the definition of "sequential compactness implies compactness".

So your interpretation of the definition is wrong. The definition merely states that a set is sequentially compact if whenever you have a sequence in $K$, it has a convergent subsequence whose limit is also in $K$.

What you have interpreted, however, is that a sequentially compact metric space is in fact compact. And as mentioned before, the other direction is also true: a compact metric space is in fact sequentially compact.

To see why, note that if $(x_n)$ is a sequence of elements from $K$, a compact subset of $X$, and assume towards contradiction that $(x_n)$ does not have a convergent subsequence; then for every $x\in K$ let $U_x$ be an open set containing only finitely many of the $x_n$'s (such $U_x$ can be found, since otherwise $x$ is a limit of a subsequence). Now $\{U_x\mid x\in K\}$ is an open cover of $K$, refining it to a finite subcover gives us at least one of the open sets with infinitely many of the $x_n$'s, which is a contradiction to how we chose the $U_x$'s.


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