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MAT 325: Topology
Professor Zoltan Szabo
Problem Set 2
Rik Sengupta
rsengupt@princeton.edu
February 7, 2010
1.
Munkres, p. 101, problem 13
Show that
X
is Hausdorﬀ if and only if the
diagonal
Δ =
{
x
×
x
:
x
∈
X
}
is closed in
X
×
X
.
Solution.
We will show both directions.
(=
⇒
:) Suppose
X
is Hausdorﬀ. We want to show that the diagonal is closed, i.e. in other
words, the complement of the diagonal,
Δ =
X
×
X

Δ, is open. Take any point (
x,y
) in
Δ.
Then, by deﬁnition,
x
6
=
y
, and hence there are open sets
U
and
V
in
X
, containing
X
and
y
respectively (because
X
is Hausdorﬀ by assumption). So, by the deﬁnition of the product
topology,
U
×
V
is an open subset of
X
×
X
. Furthermore, obviously,
U
×
V
⊂
Δ. For,
otherwise, there is some point in
U
×
V
that is in Δ. This point must be of the form (
x,x
),
but that means
x
∈
U
and
x
∈
V
, and then
U
and
V
are forced to be not disjoint, which is a
contradiction. Hence it follows that given any point (
x,y
)
∈
Δ, we can always ﬁnd some open
neighborhood
U
×
V
of that point which is completely contained in
Δ, i.e.
Δ is open. Hence,
Δ is closed.
(
⇐
=:) Suppose Δ is closed. This simply means
Δ is open. Let
x
and
y
be two distinct elements
of
X
. Then, (
x,y
)
∈
Δ by deﬁnition. Furthermore,
Δ is open, and so there is a basis open
set
U
×
V
⊂
Δ which contains (
x,y
). Now, note that
U
and
V
are open by deﬁnition of the
product topology. Furthermore, they are disjoint subsets of
X
, containing
x
and
y
respectively.
Therefore, for any unequal
x,y
∈
X
, we can ﬁnd disjoint open neighborhoods contained in
X
,
containing
x
and
y
respectively. Therefore,
X
is Hausdorﬀ.
Therefore,
X
is Hausdorﬀ if and only if the
diagonal
Δ =
{
x
×
x
:
x
∈
X
}
is closed in
X
×
X
.
±
2.
Munkres, p. 101, problem 16
Consider the ﬁve topologies on
R
given in Exercise 7 of
§
13.
(a) Determine the closure of the set
K
=
{
1
/n
:
n
∈
Z
+
}
under each of these topologies.
(b) Which of these topologies satisfy the Hausdorﬀ axiom? The
T
1
axiom?
Solution.
We will divide this solution into two parts.
Preliminary Comments.
First of all, notice that in the ﬁnite complement topology, the
sequence (
x
n
) = 1
/n
converges to all points in
R
. This is because (
x
n
) is a sequence with
inﬁnitely many distinct points. Therefore, any open set in the ﬁnite complement topology will
contain all but ﬁnitely many points of (
x
n
). It must follow, then, that (
x
n
) converges to every
point in
R
. Formally, take any
r
∈
R
, and any open set
U
r
(in the ﬁnite complement topology)
containing
r
. Since
U
r
is not empty, it follows that
R

U
r
is a ﬁnite set, by deﬁnition. Then,
deﬁne
1
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=
1
min
{
(
{
x
n
} ∩
(
R

U
r
))
∪ {
1
}}
.
Then, it is easy to see that for all
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 Fall '09
 ZoltánSzabó
 Topology

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