定义(网):
令
为一个拓扑空间,令
为一个有向集。网 是一个族
.
就像滤子一样,网是序列的模拟,它们被用来将原本只对“良好”空间成立的定理推广到一般拓扑空间的设置中。缺点是,就像滤子一样,涉及网的定理通常使用选择公理。使用网还是滤子取决于个人喜好,以及选择在给定情况下使用最少选择公理的工具。
Proof: Suppose first that
is closed, and that
is a convergent net in
to a limit
. Suppose
. Then for all
, we find
so that at least
, which implies that
, and
since
,
. Since a set is closed if and only if it contains its boundary, we conclude that
, a contradiction. Suppose then that
contains the limit of all its convergent nets. Then let
, and note that
is a directed set, ordered by inclusion. For each
, choose
by definition of
, and observe that
forms a net convergent to
. Thus,
, and since
was arbitrary,
and
is closed. 
证明: 首先假设
是连续的,并假设
收敛于
。然后令
为
的任何开邻域,并设
,由于
的连续性,它是开的。然后选取
使得对于所有
我们有
;对于这样的
我们将有
,根据原像的定义。
假设现在
具有这样的性质:如果
网收敛于
,那么
网收敛于
。然后假设
是闭集。如果我们证明
是闭集,我们就证明了该定理。 我们只需要 证明
包含其所有网的极限。所以假设
是
中的一个网,它收敛于某个
。那么
,因此
,所以
确实是闭集。 
Proof: Suppose first that
is compact, and let
be a net in
. Suppose that
does not have a convergent subnet. Then for each
, we find an open neighbourhood
of
and a
such that for all
, we have
; for otherwise, if
is a point so that for all open neighbourhoods
and
there exists
so that
, then we note that the set of all
(that is, the sets of all open neighbourhoods of
) forms a directed set when ordered inversely by inclusion (which shall henceforth be denoted by
). Note that we may take the product order on
, and then we may define an order morphism
by sending
. Consider the subset
so that
. We define a subnet of
as thus: Define
by
, and the order morphism induced by
, which is final since for any
, we find
so that
, when
is arbitrary. Now
converges to
, a contradiction. But then we apply compactness to the
, so that we gain points
so that
covers
. Then we choose an upper bound
of
by successively choosing an upper bound of
, then an upper bound of that upper bound and
and so on. Then for
,
is not contained in any of the
, so that in particular
, a contradiction. Conversely, if every net contains a convergent subnet, let
be an open cover of
. Suppose that it doesn't admit any finite subcover, and consider the set of all finite subfamilies of
ordered by inclusion, which is a directed set. Then define a net indexed over that set by choosing, for each finite index set
, the element
. By assumption, upon defining
to be the set of all finite subfamilies of
, we find that
has a convergent subnet
. Let
be the point to which this subnet converges, and since
is a cover, pick
so that
. Note that
is a finite subfamily of
, so that by finality of the order homomorphism
we find
so that
contains
. On the other hand, since
converges to
, we find
so that for
we have
. Finally, since
is directed, let
be an upper bound for
and
, then
contains
, but then
is not contained in
, a contradiction. 
定义(序列):
令
为一个拓扑空间。一个 **序列** 是一个以自然数的定向集为指标的网。
为了表示序列
收敛到点
,我们将使用有用的记号
.
命题(可数可列空间中用序列刻画闭集):
令
为一个可数可列拓扑空间。则子集
是闭集当且仅当它是序列闭的。
(关于可数选择公理的条件)
证明: 首先假设
是闭集。那么,
中任何收敛序列的极限都一定包含在
中,因为序列是一种网,而且 网的极限对应命题在没有选择公理的情况下也是成立的。现在假设
是序闭集。令
; 我们要证明
。 为邻域滤子
选择一个可数基
。 根据可数选择公理,对每个
选择
。 那么序列
收敛到
,因此
。 
命题(第一可数空间中连续性的序列判据):
设
为第一可数空间,
为拓扑空间,
为函数。则
连续当且仅当它在序列上连续。
(关于可数选择公理的条件)
证明: 若
连续,那么当然
只要
,因为
的邻域
会导致
的邻域
,
将会落在其中,当
足够大时,然后由原像的定义
。
如果
是逐点连续的,设
是闭集,并设
;我们要证明
是闭集。设
是
中的一个序列。根据第一可数空间中序列闭包的刻画,只需证明
是序列闭合的。因此,设
是
中的一个序列,它收敛于某个
。那么
,因此
,因此
。 