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Hendrik Süß
Topology
Quizzes
Hausdorff Spaces
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Mathematics
Hendrik Süß
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2004-2019
Hausdorff Spaces
Question
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There is not limit to the number of attempts for each question. However, in order to make the best use of this quiz, you should give youself sufficient time to think about every question carefully before submitting an answer. You should not only guess the answer, but also think about an argument that supports your choice. You will then be able to compare your justification with the argument in the provided model solution.
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Which of the following topologies on
ℝ
are Hausdorff?
(a)
The co-finite topology;
(b)
all subsets of the form
(
a
,
∞
)
and
∅
and
ℝ
;
(c)
all subsets
U
⊂
ℝ
such that
0
∈
U
and
∅
;
(d)
all subsets
U
⊂
ℝ
such that
0
∉
U
and
ℝ
.
There is at least one mistake.
For example, choice (a) should be
False
.
Any two non-empty open subsets have the form
U
1
=
ℝ
∖
F
1
and
U
2
=
ℝ
∖
F
2
with
F
1
,
F
2
being finite. Then
U
1
∩
U
2
=
ℝ
∖
(
F
1
∪
F
2
)
. But
(
F
1
∪
F
2
)
is finite and
ℝ
is infinite. Hence,
U
1
∩
U
2
≠
∅
.
There is at least one mistake.
For example, choice (b) should be
False
.
Again any two non-empty open subsets have the form
U
1
=
(
a
1
,
∞
)
and
U
2
=
(
a
2
,
∞
)
(where
a
1
,
a
2
∈
ℝ
∪
{
−
∞
}
). Then
U
1
∩
U
2
=
(
inf
{
a
1
,
a
2
}
,
∞
)
≠
e
m
p
t
y
s
e
t
There is at least one mistake.
For example, choice (c) should be
False
.
Obviously, we have
0
∈
U
1
∩
U
2
for any two non-empty open subsets.
There is at least one mistake.
For example, choice (d) should be
False
.
The only open subset containing
0
is
U
=
ℝ
. Now, for every open neighbourhood
V
of another point
x
∈
ℝ
we have
U
∩
V
=
V
≠
∅
.
Correct!
False
Any two non-empty open subsets have the form
U
1
=
ℝ
∖
F
1
and
U
2
=
ℝ
∖
F
2
with
F
1
,
F
2
being finite. Then
U
1
∩
U
2
=
ℝ
∖
(
F
1
∪
F
2
)
. But
(
F
1
∪
F
2
)
is finite and
ℝ
is infinite. Hence,
U
1
∩
U
2
≠
∅
.
False
Again any two non-empty open subsets have the form
U
1
=
(
a
1
,
∞
)
and
U
2
=
(
a
2
,
∞
)
(where
a
1
,
a
2
∈
ℝ
∪
{
−
∞
}
). Then
U
1
∩
U
2
=
(
inf
{
a
1
,
a
2
}
,
∞
)
≠
e
m
p
t
y
s
e
t
False
Obviously, we have
0
∈
U
1
∩
U
2
for any two non-empty open subsets.
False
The only open subset containing
0
is
U
=
ℝ
. Now, for every open neighbourhood
V
of another point
x
∈
ℝ
we have
U
∩
V
=
V
≠
∅
.
Consder a continuous map
f
:
X
→
Y
. Which of the following statements are true? Think carefully about a justification for your answer.
(a)
If
f
is surjective and
X
is Hausdorff, then
Y
is Hausdorff.
(b)
If
f
is injective and
Y
is Hausdorff, then
X
is Hausdorff.
(c)
If
f
is bijective and
Y
is Hausdorff, then
X
is Hausdorff.
(d)
If
f
is bijective and
X
is Hausdorff, then
Y
is Hausdorff.
There is at least one mistake.
For example, choice (a) should be
False
.
Remark 4.9 gives a counterexample.
There is at least one mistake.
For example, choice (b) should be
True
.
Indeed, for
x
1
,
x
2
∈
X
we have
f
(
x
1
)
≠
f
(
x
2
)
. Hence, there are disjoint open subsets
V
1
,
V
2
⊂
Y
containing
f
(
x
1
)
and
f
(
x
2
)
, respectively. Then
f
−
1
(
V
1
)
and
f
−
1
(
V
2
)
are open since
f
is continuous, disjoint and contain
x
1
and
x
2
, respectively.
There is at least one mistake.
For example, choice (c) should be
True
.
as for injectivity.
There is at least one mistake.
For example, choice (d) should be
False
.
Take
f
=
id
ℝ
:
ℝ
→
ℝ
, with the usual topology on the domain and the indiscrete topology on the codomain.
Correct!
False
Remark 4.9 gives a counterexample.
True
Indeed, for
x
1
,
x
2
∈
X
we have
f
(
x
1
)
≠
f
(
x
2
)
. Hence, there are disjoint open subsets
V
1
,
V
2
⊂
Y
containing
f
(
x
1
)
and
f
(
x
2
)
, respectively. Then
f
−
1
(
V
1
)
and
f
−
1
(
V
2
)
are open since
f
is continuous, disjoint and contain
x
1
and
x
2
, respectively.
True
as for injectivity.
False
Take
f
=
id
ℝ
:
ℝ
→
ℝ
, with the usual topology on the domain and the indiscrete topology on the codomain.
Which of the following quotient spaces are Hausdorff?
(a)
ℝ
∕
(
−
1
,
1
)
(b)
ℝ
∕
[
−
1
,
1
]
(c)
ℝ
∕
{
−
1
,
1
}
There is at least one mistake.
For example, choice (a) should be
False
.
Note, that
A
=
(
−
1
,
1
)
∈
ℝ
∕
(
−
1
,
1
)
is a single point in the identification space. The preimage
q
−
1
(
{
A
}
)
=
(
−
1
,
1
)
is not closed in
ℝ
. Hence,
{
A
}
is not closed in
ℝ
∕
(
−
1
,
1
)
, but by Proposition 4.6 singletons are closed in Hausdorff spaces. By the same argument
X
∕
A
is never Hausdorff if
A
⊂
X
is not closed.
There is at least one mistake.
For example, choice (b) should be
True
.
Consider the continuous surjection
f
:
ℝ
→
ℝ
given by
f
(
t
)
=
t
−
1
t
≥
1
,
t
+
1
t
≤
−
1
,
0
t
∈
[
−
1
,
1
]
.
This induces a continuous bijection (which is actually even a homeomorphism)
ℝ
∕
[
−
1
,
1
]
→
ℝ
. Hence, by the previous question
ℝ
∕
[
−
1
,
1
]
is Hausdorff.
Figure 1:
A schematic sketch of the map
There is at least one mistake.
For example, choice (c) should be
True
.
Consider
X
=
{
(
x
,
y
)
∣
y
=
1
or
x
2
+
y
2
=
1
}
⊂
ℝ
2
. Then the continuous surjection
f
:
ℝ
→
X
,
f
(
t
)
=
(
t
−
1
,
1
)
t
≥
1
,
(
t
+
1
,
1
)
t
≤
−
1
,
(
sin
(
2
π
t
)
,
cos
(
2
π
t
)
)
t
∈
[
−
1
,
1
]
.
Induces a continuous bijection
ℝ
∕
{
−
1
,
1
}
→
X
(which is actually a homeomorphism). Then the Hausdorff property follows by the previous question.
Figure 2:
A schematic sketch of the map
Correct!
False
Note, that
A
=
(
−
1
,
1
)
∈
ℝ
∕
(
−
1
,
1
)
is a single point in the identification space. The preimage
q
−
1
(
{
A
}
)
=
(
−
1
,
1
)
is not closed in
ℝ
. Hence,
{
A
}
is not closed in
ℝ
∕
(
−
1
,
1
)
, but by Proposition 4.6 singletons are closed in Hausdorff spaces. By the same argument
X
∕
A
is never Hausdorff if
A
⊂
X
is not closed.
True
Consider the continuous surjection
f
:
ℝ
→
ℝ
given by
f
(
t
)
=
t
−
1
t
≥
1
,
t
+
1
t
≤
−
1
,
0
t
∈
[
−
1
,
1
]
.
This induces a continuous bijection (which is actually even a homeomorphism)
ℝ
∕
[
−
1
,
1
]
→
ℝ
. Hence, by the previous question
ℝ
∕
[
−
1
,
1
]
is Hausdorff.
Figure 1:
A schematic sketch of the map
True
Consider
X
=
{
(
x
,
y
)
∣
y
=
1
or
x
2
+
y
2
=
1
}
⊂
ℝ
2
. Then the continuous surjection
f
:
ℝ
→
X
,
f
(
t
)
=
(
t
−
1
,
1
)
t
≥
1
,
(
t
+
1
,
1
)
t
≤
−
1
,
(
sin
(
2
π
t
)
,
cos
(
2
π
t
)
)
t
∈
[
−
1
,
1
]
.
Induces a continuous bijection
ℝ
∕
{
−
1
,
1
}
→
X
(which is actually a homeomorphism). Then the Hausdorff property follows by the previous question.
Figure 2:
A schematic sketch of the map
For which of the following equivalence relations is the quotient space
(
ℝ
×
{
−
1
,
1
}
)
∕
∼
Hausdorff?
(a)
The equivalence relation
∼
induced by
(
x
,
1
)
∼
(
x
,
−
1
)
.
(b)
The equivalence relation
∼
induced by
(
0
,
1
)
∼
(
0
,
−
1
)
.
(c)
The equivalence relation
∼
induced by
(
x
,
1
)
∼
(
x
,
−
1
)
for
x
≠
0
.
(d)
The equivalence relation
∼
induced by
(
x
,
1
)
∼
(
x
,
−
1
)
for
x
∈
(
−
1
,
1
)
.
(e)
The equivalence relation
∼
induced by
(
x
,
1
)
∼
(
x
,
−
1
)
for
x
∈
[
−
1
,
1
]
.
There is at least one mistake.
For example, choice (a) should be
True
.
This space can be easily seen to be homeomorphic to
ℝ
. Hence, it is Hausdorff. The homeomorphism is induced by the projection to the first factor.
There is at least one mistake.
For example, choice (b) should be
True
.
Consider
X
=
{
(
x
,
y
)
∣
x
=
−
y
or
x
=
y
}
⊂
ℝ
2
. The continuous surjection
f
:
ℝ
×
{
−
1
,
1
}
→
X
,
(
x
,
a
)
↦
(
x
,
a
x
)
.
induces a continuous bijection
(
ℝ
×
{
−
1
,
1
}
)
∕
∼
→
X
(which is actually a homeomorphism). By Question
2
the Hausdorff property follows.
Figure 3:
A schematic sketch of the map
There is at least one mistake.
For example, choice (c) should be
False
.
The
q
(
(
0
,
1
)
)
and
q
(
(
0
,
−
1
)
)
cannot be separated by open subset. The proof is very similar to Remark 4.9 in the lecture notes.
There is at least one mistake.
For example, choice (d) should be
False
.
The equivalence classes
[
(
1
,
1
)
]
and
[
(
1
,
−
1
)
]
(here
(
1
,
1
)
and
(
1
,
−
1
)
denote the pairs not the open intervals) can not be separated by open subsets. The proof is again similar to Remark 4.9 in the lecture notes.
There is at least one mistake.
For example, choice (e) should be
True
.
Consider
X
=
{
−
1
}
×
ℝ
∪
{
1
}
×
ℝ
∪
[
−
1
,
1
]
×
{
0
}
⊂
ℝ
2
.
The continuous surjection
f
:
ℝ
×
{
−
1
,
1
}
→
X
given by
f
(
x
,
a
)
=
(
1
,
a
(
x
−
1
)
)
x
≥
1
,
(
−
1
,
a
(
x
−
1
)
)
x
≤
−
1
,
(
x
,
0
)
x
∈
[
−
1
,
1
]
.
induces a continuous bijection
(
ℝ
×
{
−
1
,
1
}
)
∕
∼
→
X
(which is actually a homeomorphism). Again, the Hausdorff property follows from Question
2
.
Figure 4:
A schematic sketch of the map
Correct!
True
This space can be easily seen to be homeomorphic to
ℝ
. Hence, it is Hausdorff. The homeomorphism is induced by the projection to the first factor.
True
Consider
X
=
{
(
x
,
y
)
∣
x
=
−
y
or
x
=
y
}
⊂
ℝ
2
. The continuous surjection
f
:
ℝ
×
{
−
1
,
1
}
→
X
,
(
x
,
a
)
↦
(
x
,
a
x
)
.
induces a continuous bijection
(
ℝ
×
{
−
1
,
1
}
)
∕
∼
→
X
(which is actually a homeomorphism). By Question
2
the Hausdorff property follows.
Figure 3:
A schematic sketch of the map
False
The
q
(
(
0
,
1
)
)
and
q
(
(
0
,
−
1
)
)
cannot be separated by open subset. The proof is very similar to Remark 4.9 in the lecture notes.
False
The equivalence classes
[
(
1
,
1
)
]
and
[
(
1
,
−
1
)
]
(here
(
1
,
1
)
and
(
1
,
−
1
)
denote the pairs not the open intervals) can not be separated by open subsets. The proof is again similar to Remark 4.9 in the lecture notes.
True
Consider
X
=
{
−
1
}
×
ℝ
∪
{
1
}
×
ℝ
∪
[
−
1
,
1
]
×
{
0
}
⊂
ℝ
2
.
The continuous surjection
f
:
ℝ
×
{
−
1
,
1
}
→
X
given by
f
(
x
,
a
)
=
(
1
,
a
(
x
−
1
)
)
x
≥
1
,
(
−
1
,
a
(
x
−
1
)
)
x
≤
−
1
,
(
x
,
0
)
x
∈
[
−
1
,
1
]
.
induces a continuous bijection
(
ℝ
×
{
−
1
,
1
}
)
∕
∼
→
X
(which is actually a homeomorphism). Again, the Hausdorff property follows from Question
2
.
Figure 4:
A schematic sketch of the map