Subspaces and Product Spaces

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

Given a topological space

X

and a subset

i : Y \subset X

with subspace topology. Which implications are true?

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

The inclusion

i

is continuous with respect to the subspace topology. Hence, the composition with

f

is continuous.

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

The composition

f \circ i

cannot “see” discontinuities of

f

outside of

Y

. A concrete counterexample is given by

Y = {0} \subset ℝ = X

and

f : ℝ \to {0, 1}

given by

f (x) \{\begin{matrix} 1 & x \geq 0 \\ 0 & x < 0 . \end{matrix}

Then

f \circ i

is continuous (e.g. since

Y

is discrete), but

f

is not.

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

This is one implication of the equivalence of the universal propery of the subspace topology.

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

This is one implication of the equivalence of the universal propery of the subspace topology.

Correct!

True The inclusion $i$ is continuous with respect to the subspace topology. Hence, the composition with $f$ is continuous.
False The composition $f \circ i$ cannot “see” discontinuities of $f$ outside of $Y$ . A concrete counterexample is given by $Y = {0} \subset ℝ = X$ and $f : ℝ \to {0, 1}$ given by $f (x) \{\begin{matrix} 1 & x \geq 0 \\ 0 & x < 0 . \end{matrix}$ Then $f \circ i$ is continuous (e.g. since $Y$ is discrete), but $f$ is not.
True This is one implication of the equivalence of the universal propery of the subspace topology.
True This is one implication of the equivalence of the universal propery of the subspace topology.

Consider subsets

Y \subset X \subset ℝ^{n}

. Which of the following statements are true?

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

This follows from the fact that

B_{ϵ}^{Y} (y_{0}) = B_{ϵ}^{X} (y_{0}) \cap Y

,i.e. if

B_{ϵ}^{X} (y_{0}) \subset U

then

B_{ϵ}^{Y} (y_{0}) \subset U \cap Y

. Hence, if

U

is open in the usual topology so is

U \cap Y

. For other implication assume that

V \subset Y

is open in the usual topology. It follows, that

V

can be obtained as a union of

ϵ

-balls

B_{ϵ_{α}}^{Y} (y_{α})

for

α \in A

(as they form a basis of the usual topology). Then

U = B_{ϵ_{α}}^{X} (x_{α})

is open in

X

and

V = U \cap Y

.

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

Recall, that in the discrete topology all subsets are open. Hence, every open subset

U \cap Y

of the subspace topology is also open in the discrete topologgy of

Y

. For the other direction, note, that every subset

V \subset Y

is also a subset of

X

and as such open in the discrete topology of

X

. Hence

V = V \cap Y

is open in the subspace topology of

Y

.

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

Indeed, we may consider the only two open subsets of the indiscrete topology on

X

and consider the induced open subsets of the subspace topology on

Y

. We obtain

\emptyset \cap Y = \emptyset

and

X \cap Y = Y

, which are exactly the open subsets of the indiscrete topology on

Y

.

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

Consider the closed subset of the co-finite topology on

X

. These are

X

and all finite subsets. Now,

X \cap Y = Y

and the intersection of every finite subset with

Y

is still finite. Hence, the closed subsets of the subspace topology are also closed in the co-finite topology. For the converse note, that every finite subset

V

of

Y

is also a finite subset of

X

. Hence, every proper closed subset of

Y

with co-finite topology is also closed in

X

with respect to the co-finite topology. Now, we can write

V = V \cap Y

and

V

is therefore also closed in the subspace topology.

Correct!

True This follows from the fact that $B_{ϵ}^{Y} (y_{0}) = B_{ϵ}^{X} (y_{0}) \cap Y$ ,i.e. if $B_{ϵ}^{X} (y_{0}) \subset U$ then $B_{ϵ}^{Y} (y_{0}) \subset U \cap Y$ . Hence, if $U$ is open in the usual topology so is $U \cap Y$ . For other implication assume that $V \subset Y$ is open in the usual topology. It follows, that $V$ can be obtained as a union of $ϵ$ -balls $B_{ϵ_{α}}^{Y} (y_{α})$ for $α \in A$ (as they form a basis of the usual topology). Then $U = B_{ϵ_{α}}^{X} (x_{α})$ is open in $X$ and $V = U \cap Y$ .
True Recall, that in the discrete topology all subsets are open. Hence, every open subset $U \cap Y$ of the subspace topology is also open in the discrete topologgy of $Y$ . For the other direction, note, that every subset $V \subset Y$ is also a subset of $X$ and as such open in the discrete topology of $X$ . Hence $V = V \cap Y$ is open in the subspace topology of $Y$ .
True Indeed, we may consider the only two open subsets of the indiscrete topology on $X$ and consider the induced open subsets of the subspace topology on $Y$ . We obtain $\emptyset \cap Y = \emptyset$ and $X \cap Y = Y$ , which are exactly the open subsets of the indiscrete topology on $Y$ .
True Consider the closed subset of the co-finite topology on $X$ . These are $X$ and all finite subsets. Now, $X \cap Y = Y$ and the intersection of every finite subset with $Y$ is still finite. Hence, the closed subsets of the subspace topology are also closed in the co-finite topology. For the converse note, that every finite subset $V$ of $Y$ is also a finite subset of $X$ . Hence, every proper closed subset of $Y$ with co-finite topology is also closed in $X$ with respect to the co-finite topology. Now, we can write $V = V \cap Y$ and $V$ is therefore also closed in the subspace topology.

Consider

X = {a, b, c, d}

and

Y = {p, q, r}

with the topologies

τ_{X} = {\emptyset, X, {a}, {b, c}, {a, b, c}}

and

τ_{Y} = {\emptyset, Y, {p}, {q}, {p, q}}

, respectively. Which of the the following sets are open in the product topology?

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

Indeed,

{(b, p), (b, q), (c, p), (c, q)} = {b, c} \times {p, q}

is a product of open subsets.

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

It can be written as a union of products of open subsets:

{(a, q), (b, r), (c, r)} = ({a} \times {q}) \cup ({b, c} \times {p})

. Hence, by definition it is an element of the product topology.

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

every open subset in

X

which contains

b

also contains

c

. Hence, every product which contains an element

(b, y)

also has to contain the element

(c, y)

. Then the same holds true also for every union of products.

Correct!

True Indeed, ${(b, p), (b, q), (c, p), (c, q)} = {b, c} \times {p, q}$ is a product of open subsets.
True It can be written as a union of products of open subsets: ${(a, q), (b, r), (c, r)} = ({a} \times {q}) \cup ({b, c} \times {p})$ . Hence, by definition it is an element of the product topology.
False every open subset in $X$ which contains $b$ also contains $c$ . Hence, every product which contains an element $(b, y)$ also has to contain the element $(c, y)$ . Then the same holds true also for every union of products.

Given two topological spaces

X

and

Y

and elements

x_{0} \in X

and

y_{0} \in Y

. We consider the maps

p_{X} : X \times Y \to X; (x, y) \mapsto x,

p_{Y} : X \times Y \to Y; (x, y) \mapsto y,

i_{X} : X \to X \times Y; x \mapsto (x, y_{0}),

i_{Y} : Y \to X \times Y; y \mapsto (x_{0}, y) .

Which of the following implications hold?

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

The maps

i_{X}

and

i_{Y}

can be seen to be continuous by the universal property of the product topology. Now, the composition with a continuous map will be continuous as well.

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

Note, that the compositions with

i_{X}

and

i_{Y}

can tell us only something about the behaviour of

f

on

im i_{X} = X \times {y_{0}}

and

im i_{Y} = {x_{0}} \times Y

. Any discontinuity outside these points cannot be detected by

i_{X} \circ f

or

i_{Y} \circ f

, respectively. For a concrete counterexample consider

f : ℝ \times ℝ \to {0, 1}

given by

f (x, y) = \{\begin{matrix} 1 & x = x_{0} or y = y_{0} \\ 0 & else . \end{matrix}

The compositions above give constant maps (which are known to be continuous), but

f

cannot be continuous, since

ℝ \times ℝ

is path-connected, but

{0, 1}

is not.

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

This is one direction of the equivalence of the universal property of the product topology.

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

This is one direction of the equivalence of the universal property of the product topology.

Correct!

True The maps $i_{X}$ and $i_{Y}$ can be seen to be continuous by the universal property of the product topology. Now, the composition with a continuous map will be continuous as well.
False Note, that the compositions with $i_{X}$ and $i_{Y}$ can tell us only something about the behaviour of $f$ on $im i_{X} = X \times {y_{0}}$ and $im i_{Y} = {x_{0}} \times Y$ . Any discontinuity outside these points cannot be detected by $i_{X} \circ f$ or $i_{Y} \circ f$ , respectively. For a concrete counterexample consider $f : ℝ \times ℝ \to {0, 1}$ given by $f (x, y) = \{\begin{matrix} 1 & x = x_{0} or y = y_{0} \\ 0 & else . \end{matrix}$ The compositions above give constant maps (which are known to be continuous), but $f$ cannot be continuous, since $ℝ \times ℝ$ is path-connected, but ${0, 1}$ is not.
True This is one direction of the equivalence of the universal property of the product topology.
True This is one direction of the equivalence of the universal property of the product topology.

Consider different topology on

ℝ

Which of the following statements are true?

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

First note, that

B_{δ_{1}} (x_{1}) \times B_{δ_{2}} (x_{2})

is open in the usual topology, i.e. for ever

(x_{1}^{'}, x_{2}^{'}) \in B_{δ_{1}} (x_{1}) \times B_{δ_{2}} (x_{2})

there is an

ϵ

such that

B_{ϵ} (x_{1}^{'}, x_{2}^{'}) \subset B_{δ_{1}} (x_{1}) \times B_{δ_{2}} (x_{2})

(how to choose

ϵ

exactly?). For the other direction, consider

U

open in the usual topology. Then for every

(x_{1}, x_{2}) \in U

there is an

ϵ_{x} > 0

such that

(x_{1}, x_{2}) \in B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) \times B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) \subset B_{ϵ} (x_{1}, x_{2}) \subset U .

But then

U = ⋃_{(x_{1}, x_{2}) \in U} B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) \times B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) .

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

Every singleton

{x} \subset ℝ

is open in the discrete topology, if follows that every singleton

{(x, y)} = {x} \times {y}

is open in the product topology. But every subset of

ℝ \times ℝ

is a union of singletons. Hence, the discrete topology on

ℝ^{2}

is contained in the product topology. The other direction is trivial.

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

A basis of the product topology is given by

ℬ = {ℝ \times ℝ, \emptyset = \emptyset \times \emptyset = \emptyset \times ℝ = ℝ \times \emptyset}

, but this is already equal to the indiscrete topology, hence, closed under taking unions. Therefore the product topology coincides with the indiscrete topology.

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

Take

U = ℝ ∖ {0}

and

V = ℝ

. Then

U \times V

is open in product topology, but

ℝ^{2} ∖ U \times V = {0} \times ℝ

, which is not finite!

Correct!

True First note, that $B_{δ_{1}} (x_{1}) \times B_{δ_{2}} (x_{2})$ is open in the usual topology, i.e. for ever $(x_{1}^{'}, x_{2}^{'}) \in B_{δ_{1}} (x_{1}) \times B_{δ_{2}} (x_{2})$ there is an $ϵ$ such that $B_{ϵ} (x_{1}^{'}, x_{2}^{'}) \subset B_{δ_{1}} (x_{1}) \times B_{δ_{2}} (x_{2})$ (how to choose $ϵ$ exactly?). For the other direction, consider $U$ open in the usual topology. Then for every $(x_{1}, x_{2}) \in U$ there is an $ϵ_{x} > 0$ such that $(x_{1}, x_{2}) \in B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) \times B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) \subset B_{ϵ} (x_{1}, x_{2}) \subset U .$ But then $U = ⋃_{(x_{1}, x_{2}) \in U} B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) \times B_{\frac{1}{\sqrt{2}} ϵ_{x}} (x_{1}) .$
True Every singleton ${x} \subset ℝ$ is open in the discrete topology, if follows that every singleton ${(x, y)} = {x} \times {y}$ is open in the product topology. But every subset of $ℝ \times ℝ$ is a union of singletons. Hence, the discrete topology on $ℝ^{2}$ is contained in the product topology. The other direction is trivial.
True A basis of the product topology is given by $ℬ = {ℝ \times ℝ, \emptyset = \emptyset \times \emptyset = \emptyset \times ℝ = ℝ \times \emptyset}$ , but this is already equal to the indiscrete topology, hence, closed under taking unions. Therefore the product topology coincides with the indiscrete topology.
False Take $U = ℝ ∖ {0}$ and $V = ℝ$ . Then $U \times V$ is open in product topology, but $ℝ^{2} ∖ U \times V = {0} \times ℝ$ , which is not finite!

★	right first attempt
✓	right
✗	wrong

	(a)	$f : X \to Z$ is continuous $\Rightarrow$ $f \circ i : Y \to Z$ is continuous
	(b)	$f : X \to Z$ is continuous $\Leftarrow$ $f \circ i : Y \to Z$ is continuous?
	(c)	$g : W \to Y$ is continuous $\Rightarrow$ $i \circ g : W \to X$ is continuous
	(d)	$g : W \to Y$ is continuous $\Leftarrow$ $i \circ g : W \to X$ is continuous?

	(a)	The subspace topology on $Y$ with respect to the usual topology on $X$ is the usual topology on $Y$ .
	(b)	The subspace topology on $Y$ with respect to the discrete topology on $X$ is the discrete topology on $Y$ .
	(c)	The subspace topology on $Y$ with respect to the indiscrete topology on $X$ is the indiscrete topology on $Y$ .
	(d)	The subspace topology on $Y$ with respect to the co-finite topology on $X$ is the co-finite topology on $Y$ .

	(a)	${(b, p), (b, q), (c, p), (c, q)}$
	(b)	${(a, q), (b, p), (c, p)}$ ?
	(c)	${(a, p), (b, p), (c, q)}$

	(a)	$f : X \times Y \to Z$ is continuous $\Rightarrow$ $f \circ i_{x}$ and $f \circ i_{Y}$ are continuous.
	(b)	$f : X \times Y \to Z$ is continuous $\Leftarrow$ $f \circ i_{X}$ and $f \circ i_{Y}$ are continuous.
	(c)	$g : W \to X \times Y$ is continuous. $\Rightarrow$ $p_{X} \circ g$ and $p_{Y} \circ g$ are continuous
	(d)	$g : W \to X \times Y$ is continuous. $\Leftarrow$ $p_{X} \circ g$ and $p_{Y} \circ g$ are continuous

	(a)	The product topology on $ℝ \times ℝ$ with respect to the usual topology on both factors is the usual topology on $ℝ \times ℝ = ℝ^{2}$ .
	(b)	The product topology on $ℝ \times ℝ$ with respect to the discrete topology on both factors is the discrete topology on $ℝ^{2}$ .
	(c)	The product topology on $ℝ \times ℝ$ with respect to the indiscrete topology on both factors is the indiscrete topology on $ℝ^{2}$ .
	(d)	The product topology on $ℝ \times ℝ$ with respect to the cofinite topology on both factors is the cofinite topology on $ℝ^{2}$ .