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★ | right first attempt |

✓ | right |

✗ | wrong |

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.
*There is at least one mistake.*

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

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

For example, choice (d) should be True.
*Correct!*

For example, choice (a) should be True.

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 $Y=\left\{0\right\}\subset \mathbb{R}=X$
and $f:\mathbb{R}\to \left\{0,1\right\}$
given by

$$f\left(x\right)\left\{\begin{array}{cc}1\phantom{\rule{1em}{0ex}}\hfill & x\ge 0\hfill \\ 0\phantom{\rule{1em}{0ex}}\hfill & x<0.\hfill \\ \phantom{\rule{1em}{0ex}}\hfill \end{array}\right.$$

Then $f\circ i$ is continuous
(e.g. since $Y$ is
discrete), but $f$
is not. For example, choice (c) should be True.

For example, choice (d) should be True.

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

*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 $Y=\left\{0\right\}\subset \mathbb{R}=X$ and $f:\mathbb{R}\to \left\{0,1\right\}$ given by$$f\left(x\right)\left\{\begin{array}{cc}1\phantom{\rule{1em}{0ex}}\hfill & x\ge 0\hfill \\ 0\phantom{\rule{1em}{0ex}}\hfill & x<0.\hfill \\ \phantom{\rule{1em}{0ex}}\hfill \end{array}\right.$$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 $Y\subset X\subset {\mathbb{R}}^{n}$.
Which of the following statements are true?
*There is at least one mistake.*

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

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

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

For example, choice (d) should be True.
*Correct!*

For example, choice (a) should be True.

This follows
from the fact that ${B}_{\u03f5}^{Y}\left({y}_{0}\right)={B}_{\u03f5}^{X}\left({y}_{0}\right)\cap Y$,i.e.
if ${B}_{\u03f5}^{X}\left({y}_{0}\right)\subset U$ then
${B}_{\u03f5}^{Y}\left({y}_{0}\right)\subset U\cap Y$. 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 $\u03f5$-balls
${B}_{{\u03f5}_{\alpha}}^{Y}\left({y}_{\alpha}\right)$ for
$\alpha \in \mathcal{A}$
(as they form a basis of the usual topology). Then
$U={B}_{{\u03f5}_{\alpha}}^{X}\left({x}_{\alpha}\right)$ is open
in $X$ and
$V=U\cap Y$.

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 $Y$.

For example, choice (c) should be True.

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.

*True*This follows from the fact that ${B}_{\u03f5}^{Y}\left({y}_{0}\right)={B}_{\u03f5}^{X}\left({y}_{0}\right)\cap Y$,i.e. if ${B}_{\u03f5}^{X}\left({y}_{0}\right)\subset U$ then ${B}_{\u03f5}^{Y}\left({y}_{0}\right)\subset U\cap Y$. 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 $\u03f5$-balls ${B}_{{\u03f5}_{\alpha}}^{Y}\left({y}_{\alpha}\right)$ for $\alpha \in \mathcal{A}$ (as they form a basis of the usual topology). Then $U={B}_{{\u03f5}_{\alpha}}^{X}\left({x}_{\alpha}\right)$ 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 $\varnothing \cap Y=\varnothing $ 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 $X=\left\{a,b,c,d\right\}$
and $Y=\left\{p,q,r\right\}$ with the
topologies ${\tau}_{X}=\left\{\varnothing ,X,\left\{a\right\},\left\{b,c\right\},\left\{a,b,c\right\}\right\}$
and ${\tau}_{Y}=\left\{\varnothing ,Y,\left\{p\right\},\left\{q\right\},\left\{p,q\right\}\right\}$,
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.
*There is at least one mistake.*

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

For example, choice (c) should be False.
*Correct!*

For example, choice (a) should be True.

For example, choice (b) should be True.

It can be written as a union of products of open subsets:
$\left\{\left(a,q\right),\left(b,r\right),\left(c,r\right)\right\}=\left(\left\{a\right\}\times \left\{q\right\}\right)\cup \left(\left\{b,c\right\}\times \left\{p\right\}\right)$.
Hence, by definition it is an element of the product topology.

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
$\left(b,y\right)$ also has to contain
the element $\left(c,y\right)$.
Then the same holds true also for every union of products.

*True*Indeed, $\left\{\left(b,p\right),\left(b,q\right),\left(c,p\right),\left(c,q\right)\right\}=\left\{b,c\right\}\times \left\{p,q\right\}$ is a product of open subsets.*True*It can be written as a union of products of open subsets: $\left\{\left(a,q\right),\left(b,r\right),\left(c,r\right)\right\}=\left(\left\{a\right\}\times \left\{q\right\}\right)\cup \left(\left\{b,c\right\}\times \left\{p\right\}\right)$. 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 $\left(b,y\right)$ also has to contain the element $\left(c,y\right)$. 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
*There is at least one mistake.*

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

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

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

For example, choice (d) should be True.
*Correct!*

$${p}_{X}:X\times Y\to X;\phantom{\rule{1em}{0ex}}\left(x,y\right)\mapsto x,$$

$${p}_{Y}:X\times Y\to Y;\phantom{\rule{1em}{0ex}}\left(x,y\right)\mapsto y,$$

$${i}_{X}:X\to X\times Y;\phantom{\rule{1em}{0ex}}x\mapsto \left(x,{y}_{0}\right),$$

$${i}_{Y}:Y\to X\times Y;\phantom{\rule{1em}{0ex}}y\mapsto \left({x}_{0},y\right).$$

Which of the following implications hold?
For example, choice (a) should be True.

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
$im{i}_{X}=X\times \left\{{y}_{0}\right\}$ and
$im{i}_{Y}=\left\{{x}_{0}\right\}\times Y$.
Any discontinuity outside these points cannot be detected by
${i}_{X}\circ f$ or
${i}_{Y}\circ f$,
respectively. For a concrete counterexample consider
$f:\mathbb{R}\times \mathbb{R}\to \left\{0,1\right\}$ given
by

$$f\left(x,y\right)=\left\{\begin{array}{cc}1\phantom{\rule{1em}{0ex}}\hfill & x={x}_{0}\text{or}y={y}_{0}\hfill \\ 0\phantom{\rule{1em}{0ex}}\hfill & \text{else}.\hfill \end{array}\right.$$

The compositions above give constant maps (which are known to be continuous), but
$f$ cannot be
continuous, since $\mathbb{R}\times \mathbb{R}$ is
path-connected, but $\left\{0,1\right\}$
is not. For example, choice (c) should be True.

For example, choice (d) should be True.

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

*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 $im{i}_{X}=X\times \left\{{y}_{0}\right\}$ and $im{i}_{Y}=\left\{{x}_{0}\right\}\times Y$. Any discontinuity outside these points cannot be detected by ${i}_{X}\circ f$ or ${i}_{Y}\circ f$, respectively. For a concrete counterexample consider $f:\mathbb{R}\times \mathbb{R}\to \left\{0,1\right\}$ given by$$f\left(x,y\right)=\left\{\begin{array}{cc}1\phantom{\rule{1em}{0ex}}\hfill & x={x}_{0}\text{or}y={y}_{0}\hfill \\ 0\phantom{\rule{1em}{0ex}}\hfill & \text{else}.\hfill \end{array}\right.$$The compositions above give constant maps (which are known to be continuous), but $f$ cannot be continuous, since $\mathbb{R}\times \mathbb{R}$ is path-connected, but $\left\{0,1\right\}$ 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 $\mathbb{R}$
Which of the following statements are true?
*There is at least one mistake.*

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

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

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

For example, choice (d) should be False.
*Correct!*

For example, choice (a) should be True.

First note, that
${B}_{{\delta}_{1}}\left({x}_{1}\right)\times {B}_{{\delta}_{2}}\left({x}_{2}\right)$ is open in the usual
topology, i.e. for ever $\left({x}_{1}^{\prime},{x}_{2}^{\prime}\right)\in {B}_{{\delta}_{1}}\left({x}_{1}\right)\times {B}_{{\delta}_{2}}\left({x}_{2}\right)$
there is an $\u03f5$ such
that ${B}_{\u03f5}\left({x}_{1}^{\prime},{x}_{2}^{\prime}\right)\subset {B}_{{\delta}_{1}}\left({x}_{1}\right)\times {B}_{{\delta}_{2}}\left({x}_{2}\right)$ (how to choose
$\u03f5$ exactly?). For the other
direction, consider $U$ open in the
usual topology. Then for every $\left({x}_{1},{x}_{2}\right)\in U$
there is an ${\u03f5}_{x}>0$
such that

$$\left({x}_{1},{x}_{2}\right)\in {B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right)\times {B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right)\subset {B}_{\u03f5}\left({x}_{1},{x}_{2}\right)\subset U.$$

But then
$$U=\bigcup _{\left({x}_{1},{x}_{2}\right)\in U}{B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right)\times {B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right).$$

For example, choice (b) should be True.

Every
singleton $\left\{x\right\}\subset \mathbb{R}$
is open in the discrete topology, if follows that every singleton
$\left\{\left(x,y\right)\right\}=\left\{x\right\}\times \left\{y\right\}$
is open in the product topology. But every subset of
$\mathbb{R}\times \mathbb{R}$
is a union of singletons. Hence, the discrete topology on
${\mathbb{R}}^{2}$ is
contained in the product topology. The other direction is trivial.

For example, choice (c) should be True.

A basis of the product
topology is given by $\mathcal{\mathcal{B}}=\left\{\mathbb{R}\times \mathbb{R},\varnothing =\varnothing \times \varnothing =\varnothing \times \mathbb{R}=\mathbb{R}\times \varnothing \right\}$,
but this is already equal to the indiscrete topology, hence, closed under taking unions.
Therefore the product topology coincides with the indiscrete topology.

For example, choice (d) should be False.

Take
$U=\mathbb{R}\setminus \left\{0\right\}$ and
$V=\mathbb{R}$. Then
$U\times V$ is open in product
topology, but ${\mathbb{R}}^{2}\setminus U\times V=\left\{0\right\}\times \mathbb{R}$,
which is not finite!

*True*First note, that ${B}_{{\delta}_{1}}\left({x}_{1}\right)\times {B}_{{\delta}_{2}}\left({x}_{2}\right)$ is open in the usual topology, i.e. for ever $\left({x}_{1}^{\prime},{x}_{2}^{\prime}\right)\in {B}_{{\delta}_{1}}\left({x}_{1}\right)\times {B}_{{\delta}_{2}}\left({x}_{2}\right)$ there is an $\u03f5$ such that ${B}_{\u03f5}\left({x}_{1}^{\prime},{x}_{2}^{\prime}\right)\subset {B}_{{\delta}_{1}}\left({x}_{1}\right)\times {B}_{{\delta}_{2}}\left({x}_{2}\right)$ (how to choose $\u03f5$ exactly?). For the other direction, consider $U$ open in the usual topology. Then for every $\left({x}_{1},{x}_{2}\right)\in U$ there is an ${\u03f5}_{x}>0$ such that$$\left({x}_{1},{x}_{2}\right)\in {B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right)\times {B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right)\subset {B}_{\u03f5}\left({x}_{1},{x}_{2}\right)\subset U.$$But then$$U=\bigcup _{\left({x}_{1},{x}_{2}\right)\in U}{B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right)\times {B}_{\frac{1}{\sqrt{2}}{\u03f5}_{x}}\left({x}_{1}\right).$$*True*Every singleton $\left\{x\right\}\subset \mathbb{R}$ is open in the discrete topology, if follows that every singleton $\left\{\left(x,y\right)\right\}=\left\{x\right\}\times \left\{y\right\}$ is open in the product topology. But every subset of $\mathbb{R}\times \mathbb{R}$ is a union of singletons. Hence, the discrete topology on ${\mathbb{R}}^{2}$ is contained in the product topology. The other direction is trivial.*True*A basis of the product topology is given by $\mathcal{\mathcal{B}}=\left\{\mathbb{R}\times \mathbb{R},\varnothing =\varnothing \times \varnothing =\varnothing \times \mathbb{R}=\mathbb{R}\times \varnothing \right\}$, 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 $U=\mathbb{R}\setminus \left\{0\right\}$ and $V=\mathbb{R}$. Then $U\times V$ is open in product topology, but ${\mathbb{R}}^{2}\setminus U\times V=\left\{0\right\}\times \mathbb{R}$, which is not finite!