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CONVERGENT SEQUENCES IN TOPOLOGICAL SPACES
1. Definition and examples
Definition 1.1. Let (X, T ) be a topological space and xn ∈ X a sequence. We say that the sequence
xn converges to x0 ∈ X if for every open set U ⊆ X which contains x0 there exists an n0 ∈ N such
that for all n ≥ n0 the points xn lie in U .
Examples: 1. This definition of convergence agrees with the familiar definition (from calculus)
of convergence on Rn if we endow Rn with the Euclidean topology.
2. Let X = R be equipped with the partition topology TP associated to the partition
P = {[4a − 2, 4a + 2i | a ∈ Z}
n
Then the sequence xn = (−1) converges to 0. In fact it converges to any point in [−2, 2i. Show that
the sequence yn = sin n is also convergent.
3. If X is equipped with the trivial topology T = {∅, X} then any sequence in X is convergent
and its limit is any point in X.
4. Let X = R be given the finite complement topology. Convince yourself that the sequence
xn = n converges to x0 = 1. Does this still remain true if instead we let X have the countable
complement topology?
5. On X = R consider the particular point topology Tp with p = 0. Then the only sequences
converging to p = 0 are the sequences which are constant (and equal to zero) after some n0 , that is
there is some n0 such that for all n ≥ n0 we have xn = 0. Describe all sequences which converge to
x0 = 1!
2. Uniqueness of limit and Hausdorff spaces
As the examples above show, in general topological spaces limits of sequences may not be unique.
They are unique tough if X happens to have the Hausdorff property.
Definition 2.1. A topological space X is called Hausdorff if for every two points a, b ∈ X there exist
disjoint open sets Ua , Ub ⊆ X with a ∈ Ua and b ∈ Ub .
Theorem 2.2. Let X be a Hausdorff space and xn ∈ X a convergent sequence. Then the limit lim xn
n→∞
is unique.
Proof. Suppose that there are two (or more) limits, say a and b. Since X is Hausdorff, we can find
disjoint open sets Ua and Ub with a ∈ Ua and b ∈ Ub . Let na ∈ N be such that xn ∈ Ua for all n ≥ na
and nb ∈ N have the property that xn ∈ Ub for all n ≥ nb . Then for all n ≥ max{na , nb } we have that
xn ∈ Ua ∩ Ub which is a contradiction since Ua ∩ Ub = ∅.
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The converse of theorem 2.2 is not generally true. That is, there are examples of topological spaces
which are not Hausdorff even tough every of their convergent sequences has a unique limit. Here is
one such example.
Example 6: Consider X = R with the countable complement topology. Let us first check that
every convergent sequence xn ∈ X has a unique limit. Suppose not, that is suppose that lim xn = a
and lim xn = b with a 6= b. Let Ua be the open set
Ua = R − {xi | xi 6= a}
Clearly a ∈ Ua and so there must be some na ∈ N such that xn ∈ Ua for all n ≥ na . But then xn = a
for all n ≥ na since xn ∈ Ua ∩ {xi | i ∈ N} = {a}. A similar argument shows that for some nb ∈ N
all xn = b for n ≥ nb . But then xn = a and xn = b for n ≥ max{na , nb } which is impossible since
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CONVERGENT SEQUENCES IN TOPOLOGICAL SPACES
a 6= b. On the other hand, X is not Hausdorff since every two non-empty open sets have nontrivial
intersection.
There is still a broad class of topological spaces for which the converse of theorem 2.2 does hold
true. To describe this class of spaces we first need the following definition.
Definition 2.3. Let (X, T ) be a topological space and x ∈ X a point. A collection of open sets Bx is
called a basis at x if for every open set U ⊆ X containing x there is some element V ∈ Bx such that
V ⊆ U . A topological space (X, T ) is called first-countable if every point x ∈ X has a countable
basis at x.
Note that if X is second-countable then it is first-countable (but not the other way around). We
now have the following converse of theorem 2.2.
Theorem 2.4. If X is a first-countable topological space and has the property that every convergent
sequence has a unique limit then X is Hausdorff.
Proof. Given two arbitrary points a, b ∈ X we need to find two disjoint open sets of which one contains
a and the other contains b.
Let Ba = {Uia ⊆ X | i ∈ N} and Bb = {Uib ⊆ X | i ∈ N} be countable bases at a and b respectively.
We define new open sets Via and Vib as
Via = U1a ∩ U2a ∩ ... ∩ Uia
and
Vib = U1b ∩ U2b ∩ ... ∩ Uib
These sets are open sets since they are obtained as finite intersections of open sets. Furthermore,
notice that Vja ⊆ Uia and Vjb ⊆ Uib for every j ≥ i and a ∈ Via and b ∈ Vib for every i ∈ N.
If Via ∩Vib = ∅ for some i we have proved the theorem. Suppose thus that Via ∩Vib 6= ∅ for any i ∈ N.
Let xi ∈ Via ∩ Vib be an arbitrary point. This yields a sequence in X which we claim converges to a.
To see this, let U be an open set containing a. Let Una ∈ Ba be an open set such that Una ⊆ U . Then
Vja ⊆ U for every j ≥ n, in particular xj ∈ U for every j ≥ n. This means that lim xi = a. Repeating
this same argument for b shows also that lim xi = b. This is a contradiction since by assumption all
convergent sequences in X have a unique limit.
We conclude that there is some i ∈ N for which Via ∩ Vib = ∅. Since a, b ∈ X were chosen arbitrarily,
X is a Hausdorff space.
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3. Convergent sequences and closed sets
Recall that we saw that a subset A ⊆ Rn (with the Euclidean topology) is closed if and only if it
contains the limits of all its convergent sequences (this was our original definition of a closed subset
of Rn ). This characterization of closed sets is only “half-way”true in general topological spaces.
Theorem 3.1. If A ⊆ X is a closed subset of X, then A contains the limits of all its convergent
sequences.
Proof. Suppose an ∈ A is a convergent sequence with lim an = a0 . If a0 ∈
/ A then a0 ∈ X − A which is
an open set. Since an is convergent, there must be some n0 ∈ N such that an ∈ X − A for all n ≥ n0 .
This is impossible since an ∈ A and A ∩ (X − A) = ∅.
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That the converse of this theorem is not true in general topological spaces is illustrated by the
following example.
Example 7: Let X = R be equipped with the countable complement topology and let A ⊆ X be
the set A = X − {0}. Notice that A is not closed (since X − A = {0} is not open). But A contains
the limits of all of its convergent subsequences. To see this we only need to show that no sequence
xn ∈ A can converge to 0. This is easy to see since the set U = X − {x1 , x2 , ...} is an open set which
contains zero but not element of the sequence xn . Thus A contains the limits of all of its convergent
sequences.
Nonetheless, as in the previous section, there is a class of topological spaces for which the converse
of theorem 3.1 is still true. Before stating the theorem we first need the following lemma.
CONVERGENT SEQUENCES IN TOPOLOGICAL SPACES
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Lemma 3.2. Let X be a topological space and A ⊆ X a subset of X. If p ∈ Ā − A then every open
set U which contains p must intersect A.
Proof. Suppose not. Then there exists an open set U such that p ∈ U and A ∩ U = ∅. But then the
set Ā − U is a closed set and A ⊆ Ā − U ⊂ Ā. This is a contradiction because Ā is the smallest closed
set containing A.
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Theorem 3.3. Let X be a first-countable topological space and A ⊆ X a subset of X with the property
that it contains the limits of all of its convergent sequences. Then A is closed.
Proof. We will show that A is closed by exhibiting that A = Ā. Suppose not, then Ā − A is nonempty.
Let p ∈ Ā − A and let Bp = {Ui ⊆ X | i = 1, 2, 3, ...} be a countable basis at p. As in the proof of
theorem 2.4 we define a new family Vi of open sets as
Vi = U1 ∩ U2 ∩ ... ∩ Ui
The sets Vi are open and Vj ⊆ Ui for all j ≥ i. Each Vi contains p. By lemma 3.2 each set Vi ∩ A is
non-empty. Pick an arbitrary element xi ∈ Vi ∩ A. Arguing as in the proof of theorem 2.4, we find
that xi is a convergent sequence with lim xi = p. Since xi ∈ A and lim xi = p we conclude that p ∈ A.
Thus Ā = A.
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This theorem together with example 7 shows that (R, Tcc ) (where Tcc is the countable complement
topology) is not a first-countable space (and hence also not second-countable). Thus (R, Tcc ) is not
homeomorphic to (R, TE ) (where TE it the Euclidean topology on R).
4. Convergent sequences and continuous functions
The relation between convergent sequences and continuous functions in general topological spaces
is the same as it is in Euclidean space:
Theorem 4.1. Let f : X → Y be a continuous function between two topological spaces. If xi ∈ X is
a convergent sequence with lim xi = x0 then f (xi ) ∈ Y is also a convergent sequence with lim f (xi ) =
f (x0 ).
Proof. Let V ⊆ Y be an open set containing f (x0 ). Since f is continuous, the set U = f −1 (V ) ⊆ X
is an open set in X containing x0 . Since lim xi = x0 there is some n0 ∈ N such that xn ∈ U for all
n ≥ n0 . But then f (xn ) ∈ V for all n ≥ n0 showing that lim f (xi ) = f (x0 ).
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