Compactness

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.

Start Quiz

Which of the following assertions are true?

There is at least one mistake.
For example, choice (a) should be True.

This is Theorem 5.8.

There is at least one mistake.
For example, choice (b) should be False.

Consider

ℝ \subset ℝ

.

There is at least one mistake.
For example, choice (c) should be False.

Consider

{a, b} \subset {a, b}

with the indiscrete topology.

There is at least one mistake.
For example, choice (d) should be True.

This is even true for arbitrary subsets of a Hausdorff space (see Propostion 4.8)

There is at least one mistake.
For example, choice (e) should be True.

This is Proposition 5.9.

Correct!

True This is Theorem 5.8.
False Consider $ℝ \subset ℝ$ .
False Consider ${a, b} \subset {a, b}$ with the indiscrete topology.
True This is even true for arbitrary subsets of a Hausdorff space (see Propostion 4.8)
True This is Proposition 5.9.

Consider subsets

X, Y \subset Z

of some Hausdorff space. Which of the following implications do hold?

There is at least one mistake.
For example, choice (a) should be True.

This is one direction of Theorem 5.11.

There is at least one mistake.
For example, choice (b) should be True.

This is the other direction of Theorem 5.11.

There is at least one mistake.
For example, choice (c) should be True.

Take an open cover

ℱ

of

X \cup Y

. Then

ℱ

is also an open cover for

X

and

Y

, respectively. Compactness of

X

and

Y

implies the existence of finite subcovers

ℱ_{X}

and

ℱ_{Y}

of

X

and

Y

, respectively. Then

ℱ_{X} \cup ℱ_{Y} \subset ℱ

is a finite subcover for

X \cup Y

.

There is at least one mistake.
For example, choice (d) should be False.

Consider

[0, 1] = [0, 1) \cup (0, 1]

.

There is at least one mistake.
For example, choice (e) should be True.

By Theorem 5.8 the subsets

X

and

Y

are closed. Hence, the intersection

X \cap Y \subset Z

is closed as well. Then

X \cap Y \subset X

is also closed in the subspace topology on

X

. As a closed subset of a compact topological space

X \cap Y

is compact as well. Note, that the Hausdorff property is essential.

There is at least one mistake.
For example, choice (f) should be False.

Consider

[0, 1] = (- \infty, 1] \cap [0, \infty)

.

Correct!

True This is one direction of Theorem 5.11.
True This is the other direction of Theorem 5.11.
True Take an open cover $ℱ$ of $X \cup Y$ . Then $ℱ$ is also an open cover for $X$ and $Y$ , respectively. Compactness of $X$ and $Y$ implies the existence of finite subcovers $ℱ_{X}$ and $ℱ_{Y}$ of $X$ and $Y$ , respectively. Then $ℱ_{X} \cup ℱ_{Y} \subset ℱ$ is a finite subcover for $X \cup Y$ .
False Consider $[0, 1] = [0, 1) \cup (0, 1]$ .
True By Theorem 5.8 the subsets $X$ and $Y$ are closed. Hence, the intersection $X \cap Y \subset Z$ is closed as well. Then $X \cap Y \subset X$ is also closed in the subspace topology on $X$ . As a closed subset of a compact topological space $X \cap Y$ is compact as well. Note, that the Hausdorff property is essential.
False Consider $[0, 1] = (- \infty, 1] \cap [0, \infty)$ .

Assume

f : X \to Y

is a continuous map from a compact space to a Hausdorff space. Which of the following statements are necessarily true? Think carefully about a justification of your answer!

There is at least one mistake.
For example, choice (a) should be True.

This is the definition of continuity of

f

.

There is at least one mistake.
For example, choice (b) should be True.

This is equivalent to the continuity of

f

.

There is at least one mistake.
For example, choice (c) should be False.

Take

f : {0} \to ℝ

the inclusion map. Then

f ({0}) = {0}

, which is not open in

ℝ

.

There is at least one mistake.
For example, choice (d) should be True.

A closed subset

A \subset X

is compact, since

X

is compact. Then its image

f (A)

is compact, since

f

is continuous. Now, it follows that

f (A)

is closed, since

Y

is Hausdorff.

Correct!

True This is the definition of continuity of $f$ .
True This is equivalent to the continuity of $f$ .
False Take $f : {0} \to ℝ$ the inclusion map. Then $f ({0}) = {0}$ , which is not open in $ℝ$ .
True A closed subset $A \subset X$ is compact, since $X$ is compact. Then its image $f (A)$ is compact, since $f$ is continuous. Now, it follows that $f (A)$ is closed, since $Y$ is Hausdorff.

Every continuous bijection between two bounded and closed subsets of Euclidean space is a homeomorphism.

Choice (a) is Correct!

Bounded and closed subsets of Euclidean space are compact by the Heine-Borel Theorem. As subspaces of a Hausdorff space they are also Hausdorff. Hence, every continuous bijection between those spaces is a homeomorphism.

Choice (b) is Incorrect.

Bounded and closed subsets of Euclidean space are compact by the Heine-Borel Theorem. As subspaces of a Hausdorff space they are also Hausdorff. Hence, every continuous bijection between those spaces is a homeomorphism.

Consider two topologies

τ

and

τ^{'}

on the same topological space

X

such that

τ^{'} ⊊ τ

. Which of the following assertions are true?

There is at least one mistake.
For example, choice (a) should be True.

Indeed, consider

U \in τ^{'}

. Then

{id}_{X}^{- 1} (U) = U \in τ^{'} \subset τ

.

There is at least one mistake.
For example, choice (b) should be False.

Pick any

U \in τ ∖ τ^{'}

. Then

{id}_{X}^{- 1} (U) = U \notin τ^{'}

.

Correct!

True Indeed, consider $U \in τ^{'}$ . Then ${id}_{X}^{- 1} (U) = U \in τ^{'} \subset τ$ .
False Pick any $U \in τ ∖ τ^{'}$ . Then ${id}_{X}^{- 1} (U) = U \notin τ^{'}$ .

Consider two topologies

τ

and

τ^{'}

on the same topological space

X

such that

τ^{'} \subset τ

. Which of the following assertions are true?

There is at least one mistake.
For example, choice (a) should be False.

Take

X = ℝ

and

τ

to be the discrete and

τ^{'}

to be the indiscrete topology.

There is at least one mistake.
For example, choice (b) should be True.

Given two distinct elements

x, y \in X

. If they admit two disjoint open neighbourhoods from

τ^{'}

then the same subsets are also disjoint open nieghbourhoods with respect to

τ

, since

τ^{'} \subset τ

.

There is at least one mistake.
For example, choice (c) should be True.

Given an cover

ℱ

of

X

consisting of elements from

τ^{'}

. Then this is also an open cover with respect to

τ

. Hence, by compactness of

(X, τ)

it admits a finite subcover.

There is at least one mistake.
For example, choice (d) should be False.

Take

X = ℝ

and

τ

to be the discrete and

τ^{'}

to be the indiscrete topology.

There is at least one mistake.
For example, choice (e) should be False.

Take

X = ℝ

and

τ

to be the discrete and

τ^{'}

to be the indiscrete topology.

There is at least one mistake.
For example, choice (f) should be True.

Note that

{id}_{X} : (X, τ) \to (X, τ^{'})

is a continuous bijection from a compact space to a Hausdorff space. Hence, it is a homeomorphism. Therefore

U \in τ \Leftrightarrow U = {id}_{X} (U) \in τ^{'}

.

Correct!

False Take $X = ℝ$ and $τ$ to be the discrete and $τ^{'}$ to be the indiscrete topology.
True Given two distinct elements $x, y \in X$ . If they admit two disjoint open neighbourhoods from $τ^{'}$ then the same subsets are also disjoint open nieghbourhoods with respect to $τ$ , since $τ^{'} \subset τ$ .
True Given an cover $ℱ$ of $X$ consisting of elements from $τ^{'}$ . Then this is also an open cover with respect to $τ$ . Hence, by compactness of $(X, τ)$ it admits a finite subcover.
False Take $X = ℝ$ and $τ$ to be the discrete and $τ^{'}$ to be the indiscrete topology.
False Take $X = ℝ$ and $τ$ to be the discrete and $τ^{'}$ to be the indiscrete topology.
True Note that ${id}_{X} : (X, τ) \to (X, τ^{'})$ is a continuous bijection from a compact space to a Hausdorff space. Hence, it is a homeomorphism. Therefore $U \in τ \Leftrightarrow U = {id}_{X} (U) \in τ^{'}$ .

★	right first attempt
✓	right
✗	wrong

	(a)	A compact subspace of a Hausdorff space is closed.
	(b)	A closed subspace of a Hausdorff space is compact.
	(c)	A closed subspace of a compact space is Hausdorff.
	(d)	A closed subspace of a Hausdorff space is Hausdorff.
	(e)	A closed subspace of a compact space is compact.

	(a)	$X$ and $Y$ are both compact $\Rightarrow$ $X \times Y$ is compact.
	(b)	$X$ and $Y$ are both compact $\Leftarrow$ $X \times Y$ is compact.
	(c)	$X$ and $Y$ are both compact $\Rightarrow$ $X \cup Y$ is compact.
	(d)	$X$ and $Y$ are both compact $\Leftarrow$ $X \cup Y$ is compact.
	(e)	$X$ and $Y$ are both compact $\Rightarrow$ $X \cap Y$ is compact.
	(f)	$X$ and $Y$ are both compact $\Leftarrow$ $X \cap Y$ is compact.

	(a)	The preimage of an open subset is open.
	(b)	The preimage of a closed subset is closed.
	(c)	The image of an open subset is open.
	(d)	The image of a closed subset is closed.

	(a)	${id}_{X} : (X, τ) \to (X, τ^{'})$ is continuous.
	(b)	${id}_{X} : (X, τ^{'}) \to (X, τ)$ is continuous.

	(a)	If $(X, τ)$ is Hausdorff so is $(X, τ^{'})$ .
	(b)	If $(X, τ^{'})$ is Hausdorff so is $(X, τ)$ .
	(c)	If $(X, τ)$ is compact so is $(X, τ^{'})$ .
	(d)	If $(X, τ^{'})$ is compact so is $(X, τ)$ .
	(e)	If $(X, τ^{'})$ is compact and $(X, τ)$ is Hausdorff, then $τ = τ^{'}$ .
	(f)	If $(X, τ)$ is compact and $(X, τ^{'})$ is Hausdorff, then $τ = τ^{'}$ .