❖
✖ Questions

- Howto

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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 continuous map $f:X\to Y$,
which of the following implications necessarily hold?
*There is at least one mistake.*

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

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be True.

For example, choice (d) should be False.

Consider
e.g. a constant map $f:\mathbb{R}\to \mathbb{R}$.
Then every preimage is open.

*False*Consider e.g. a constant map $\mathbb{R}\to \mathbb{R}$*False*Consider e.g. $f:\mathbb{R}\to \left\{0\right\}$. Then the image of $f\left(U\right)$ of every subset $U\subset \mathbb{R}$ is open.*True*This is actually the definition of continuity.*False*Consider e.g. a constant map $f:\mathbb{R}\to \mathbb{R}$. Then every preimage is open.

The set of singletons $\left\{\left\{x\right\}\mid x\in \mathbb{R}\right\}$ is a basis
of the
*Correct!*
*Incorrect.* *Please try again*

Every subset of $\mathbb{R}$
is the union of its singletons. Hence, the topologly generated by the singletons is
$\mathcal{P}\left(\mathbb{R}\right)$ – the discrete topology.

Every subset of $\mathbb{R}$
is the union of its singletons. Hence, the topologly generated by the singletons is
$\mathcal{P}\left(\mathbb{R}\right)$ – the
discrete topology.

Given two topologies ${\tau}_{1}$
and ${\tau}_{2}$ on a set
$X$, which of the following
sets is again a topology
*There is at least one mistake.*

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

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

For example, choice (a) should be False.

For example, choice (b) should be True.

For example, choice (c) should be False.

For example, choice (d) should be False.

$X$ and
$\varnothing $ are never
contained in $\left({\tau}_{1}\cup {\tau}_{2}\right)\setminus \left({\tau}_{1}\cap {\tau}_{2}\right)$.

*False*Consider the topologies ${\tau}_{1}=\left\{\varnothing ,\left\{a\right\},X\right\}$ and ${\tau}_{2}=\left\{\varnothing ,\left\{b\right\},X\right\}$ on $X=\left\{a,b,c\right\}$. Then ${\tau}_{1}\cup {\tau}_{2}=\left\{\varnothing ,\left\{a\right\},\left\{b\right\},X\right\}$ is not a topology.*True**False*$X$ and $\varnothing $ are never contained in ${\tau}_{1}\setminus {\tau}_{2}$*False*$X$ and $\varnothing $ are never contained in $\left({\tau}_{1}\cup {\tau}_{2}\right)\setminus \left({\tau}_{1}\cap {\tau}_{2}\right)$.

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 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 False.
*Correct!*

For example, choice (a) should be True.

For example, choice (b) should be False.

Consider
${\mathcal{\mathcal{B}}}_{1}=\left\{\left\{a\right\},\left\{b\right\},\left\{c\right\},\left\{a,b\right\}\right\}$ and
${\mathcal{\mathcal{B}}}_{2}=\left\{\left\{a\right\},\left\{b\right\},\left\{c\right\},\left\{a,c\right\}\right\}$ both are bases of the
discrete topology on $\left\{a,b,c\right\}$.

For example, choice (c) should be True.

For example, choice (d) should be False.

The intersection might even be empty: take
${\mathcal{\mathcal{B}}}_{1}$ to be the set of
open balls in ${\mathbb{R}}^{2}$
and ${\mathcal{\mathcal{B}}}_{2}$ the set of
products $I\times J$ of
open intervals $I,J\subset \mathbb{R}$.
Extra question: is the statement true for finite topological spaces?

*True**False*Consider ${\mathcal{\mathcal{B}}}_{1}=\left\{\left\{a\right\},\left\{b\right\},\left\{c\right\},\left\{a,b\right\}\right\}$ and ${\mathcal{\mathcal{B}}}_{2}=\left\{\left\{a\right\},\left\{b\right\},\left\{c\right\},\left\{a,c\right\}\right\}$ both are bases of the discrete topology on $\left\{a,b,c\right\}$.*True*If any open subset can be written as a union of elements of ${\mathcal{\mathcal{B}}}_{1}$, then obviously it can also be written as union of elements of ${\mathcal{\mathcal{B}}}_{1}\cup {\mathcal{\mathcal{B}}}_{2}$ (we can take the same elements).*False*The intersection might even be empty: take ${\mathcal{\mathcal{B}}}_{1}$ to be the set of open balls in ${\mathbb{R}}^{2}$ and ${\mathcal{\mathcal{B}}}_{2}$ the set of products $I\times J$ of open intervals $I,J\subset \mathbb{R}$. Extra question: is the statement true for finite topological spaces?

Consider the co-finite topology on $\mathbb{R}$
(i.e $\varnothing $,
and all subsets with finite complement are open). Which of the following functions
$f:\mathbb{R}\to \mathbb{R}$ are
continuous with respect to the co-finite topology on the domain and codomain?
*There is at least one mistake.*

For example, choice (a) should be False.
*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 False.

For example, choice (b) should be True.

Consider a proper closed, hence finite, subset
$V=\left\{{a}_{1},\dots ,{a}_{m}\right\}$. Assume first, that
$f$ is not constant. The polynomial
equation $f\left(x\right)={a}_{i}$ has only finitely
many solutions for every $i=1,\dots ,m$.
It follows, that ${f}^{-1}\left(V\right)$,
which is the union of these solution sets, is finite, hence closed. If
$f\equiv a$ is constant,
then either ${f}^{-1}\left(V\right)=\varnothing $
(if $a\notin V$) or
${f}^{-1}\left(V\right)=\mathbb{R}$ (if
$a\in V$). In any case it is
always an open subset.

For example, choice (c) should be True.

For example, choice (d) should be False.

${f}^{-1}\left(1\right)=\mathbb{R}\setminus \left\{0\right\}$ is a proper infinite
subset of $\mathbb{R}$, hence it
is not closed, but $\left\{1\right\}$
is finite, hence closed.

*False*The preimage of the finite, hence closed, set $\left\{0\right\}$ is given by ${f}^{-1}\left(0\right)=\mathbb{Z}\pi \u228a\mathbb{R}$, which is infinite, hence not closed.*True*Consider a proper closed, hence finite, subset $V=\left\{{a}_{1},\dots ,{a}_{m}\right\}$. Assume first, that $f$ is not constant. The polynomial equation $f\left(x\right)={a}_{i}$ has only finitely many solutions for every $i=1,\dots ,m$. It follows, that ${f}^{-1}\left(V\right)$, which is the union of these solution sets, is finite, hence closed. If $f\equiv a$ is constant, then either ${f}^{-1}\left(V\right)=\varnothing $ (if $a\notin V$) or ${f}^{-1}\left(V\right)=\mathbb{R}$ (if $a\in V$). In any case it is always an open subset.*True*Consider a proper closed, hence finite, subset $V\subset \mathbb{R}$. Then either ${f}^{-1}\left(V\right)=V$ (if $0\notin V$) or ${f}^{-1}\left(V\right)=V\cup \left\{1\right\}$ (if $0\in V$). In any case the preimage is finite, hence closed.*False*${f}^{-1}\left(1\right)=\mathbb{R}\setminus \left\{0\right\}$ is a proper infinite subset of $\mathbb{R}$, hence it is not closed, but $\left\{1\right\}$ is finite, hence closed.

Which of the following topologies on $X=\left\{a,b\right\}$
is path-connected?
*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.

For example, choice (b) should be True.

$\tau \left(t\right)=\left\{\begin{array}{cc}a\phantom{\rule{1em}{0ex}}\hfill & t=0\hfill \\ b\phantom{\rule{1em}{0ex}}\hfill & t\in \left(0,1\right]\hfill \end{array}\right.$defines a
path from $a$ to
$b$. Indeed, it is
continuous, since ${\sigma}^{-1}\left(\left\{1\right\}\right)=\left(0,1\right]$
is open in $\left[0,1\right]$.

For example, choice (c) should be True.

$\gamma \left(t\right)=\left\{\begin{array}{cc}a\phantom{\rule{1em}{0ex}}\hfill & t=\left[0,1\right)\hfill \\ b\phantom{\rule{1em}{0ex}}\hfill & t=1\hfill \end{array}\right.$defines a
path from $a$ to
$b$. Indeed, it is
continuous, since ${\gamma}^{-1}\left(\left\{0\right\}\right)=\left[0,1\right)$
is open in $\left[0,1\right]$.

For example, choice (d) should be False.

Assume there
is a path from $\delta $
from $a$ to
$b$. We may
compose $\delta $
with the map $X\to \mathbb{R}$
given by $a\mapsto 0$ and
$b\mapsto 1$, which is
continuous since $X$
is a discrete space. Hence, we obtain a continuous map
$\left[0,1\right]\to \mathbb{R}$ taking
values in $0$
an $1$,
but no values in between. A contradiction to the Intermediate Values Theorem.

*True*$\sigma \left(t\right)=\left\{\begin{array}{cc}a\phantom{\rule{1em}{0ex}}\hfill & t=0\hfill \\ b\phantom{\rule{1em}{0ex}}\hfill & t\in \left(0,1\right]\hfill \end{array}\right.$defines a path from $a$ to $b$. Indeed, it is continuous, since every map to an indiscrete space is continuous.*True*$\tau \left(t\right)=\left\{\begin{array}{cc}a\phantom{\rule{1em}{0ex}}\hfill & t=0\hfill \\ b\phantom{\rule{1em}{0ex}}\hfill & t\in \left(0,1\right]\hfill \end{array}\right.$defines a path from $a$ to $b$. Indeed, it is continuous, since ${\sigma}^{-1}\left(\left\{1\right\}\right)=\left(0,1\right]$ is open in $\left[0,1\right]$.*True*$\gamma \left(t\right)=\left\{\begin{array}{cc}a\phantom{\rule{1em}{0ex}}\hfill & t=\left[0,1\right)\hfill \\ b\phantom{\rule{1em}{0ex}}\hfill & t=1\hfill \end{array}\right.$defines a path from $a$ to $b$. Indeed, it is continuous, since ${\gamma}^{-1}\left(\left\{0\right\}\right)=\left[0,1\right)$ is open in $\left[0,1\right]$.*False*Assume there is a path from $\delta $ from $a$ to $b$. We may compose $\delta $ with the map $X\to \mathbb{R}$ given by $a\mapsto 0$ and $b\mapsto 1$, which is continuous since $X$ is a discrete space. Hence, we obtain a continuous map $\left[0,1\right]\to \mathbb{R}$ taking values in $0$ an $1$, but no values in between. A contradiction to the Intermediate Values Theorem.

Every set $\mathcal{\mathcal{B}}$ of subsets
of $X$ is the basis of
some topology on $X$.
*Choice (a) is Incorrect.*
*Choice (b) is Correct!*

Consider $X=\left\{a,b,c\right\}$.
Assume that $\mathcal{\mathcal{B}}$ is a
basis of a topology $\tau $.

- Then $X\in \tau $. Hence, $X={\bigcup}_{U\in \mathcal{\mathcal{B}}}U$, but this does not hold, e.g. for $\mathcal{\mathcal{B}}=\left\{a\right\}$.
- For $U,V\in \mathcal{\mathcal{B}}\subset \tau $ we have $U\cap V\in \tau $. Hence, for every pair $U,V\in \mathcal{\mathcal{B}}$ the intersection $U\cap V$ must be a union of subsets in $\mathcal{\mathcal{B}}$. This does not hold, e.g. for $X=\left\{a,b,c\right\}$ and $\mathcal{\mathcal{B}}=\left\{\left\{a,b\right\},\left\{b,c\right\}\right\}$.

Consider $X=\left\{a,b,c\right\}$.
Assume that $\mathcal{\mathcal{B}}$ is a
basis of a topology $\tau $.

- Then $X\in \tau $. Hence, $X={\bigcup}_{U\in \mathcal{\mathcal{B}}}U$, but this does not hold, e.g. for $\mathcal{\mathcal{B}}=\left\{\left\{a\right\}\right\}$.
- For $U,V\in \mathcal{\mathcal{B}}\subset \tau $ we have $U\cap V\in \tau $. Hence, for every pair $U,V\in \mathcal{\mathcal{B}}$ the intersection $U\cap V$ must be a union of subsets in $\mathcal{\mathcal{B}}$. This does not hold, e.g. for $\mathcal{\mathcal{B}}=\left\{\left\{a,b\right\},\left\{b,c\right\}\right\}$.