❖
✖ 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.

Consider the subset $X=\left(\left(-\infty ,0\right]\times \left\{1\right\}\right)\phantom{\rule{2.77695pt}{0ex}}\cup \phantom{\rule{2.77695pt}{0ex}}\left(\left(0,\infty \right)\times \left\{0\right\}\right)\subset {\mathbb{R}}^{2}$.

The map $f:X\to \mathbb{R}$ defined by $f\left(x,y\right)=x$ is
*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!*

The map $f:X\to \mathbb{R}$ defined by $f\left(x,y\right)=x$ is

For example, choice (a) should be True.

For example, choice (b) should be True.

For example, choice (c) should be False.

The inverse of $f$
is not continuous at $0$.
Indeed, take ${x}_{n}=1\u2215n$.
Then $\underset{n\to \infty}{lim}{x}_{n}=0$, but
${f}^{-1}\left({x}_{n}\right)=\left(1\u2215n,0\right)$ does not
converge in $X$ (it
converges in ${\mathbb{R}}^{2}$ to
$\left(0,0\right)$, but this is not
an element of $X$).

*True*Since it is the restriction of a continuous map (the projection ${\mathbb{R}}^{2}\to \mathbb{R}$).*True*Indeed, an inverse is given by ${f}^{-1}\left(x\right)=\left\{\begin{array}{cc}\left(x,1\right)\phantom{\rule{1em}{0ex}}\hfill & x\le 0\hfill \\ \left(x,0\right)\phantom{\rule{1em}{0ex}}\hfill & x>0.\hfill \end{array}\right.$*False*The inverse of $f$ is not continuous at $0$. Indeed, take ${x}_{n}=1\u2215n$. Then $\underset{n\to \infty}{lim}{x}_{n}=0$, but ${f}^{-1}\left({x}_{n}\right)=\left(1\u2215n,0\right)$ does not converge in $X$ (it converges in ${\mathbb{R}}^{2}$ to $\left(0,0\right)$, but this is not an element of $X$).

Which of the following properties of subsets of Euclidean space are topological properties?
*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 False.
*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.

For example, choice (c) should be False.

For example, choice (d) should be True.

The argument is the same as for path-connectedness: a path
between two points induces a path between their images under a continuous map.

*True*A homeomorphism is a bijection. Hence, given a bijection of one subset to $\mathbb{N}$ one obtains a bijection of a homeomorphic subset to $\mathbb{N}$ via composition.*False*Consider the interval $\left[-1,1\right]$ and the upper semi-circle. The first is convex, the second is not.*False*Take $\mathbb{R}$ and $\left(-1,1\right)$. The first is complete, the second is not.*True*The argument is the same as for path-connectedness: a path between two points induces a path between their images under a continuous map.

Consider the subset $X=\left\{\left(x,y\right)\in {\mathbb{R}}^{2}\mid xy=1\right\}\subset {\mathbb{R}}^{2}$.

The map $f:X\to \mathbb{R}\setminus \left\{0\right\}$ defined by $f\left(x,y\right)=x$ is a homeomorphism.
*Choice (a) is Correct!*
*Choice (b) is Incorrect.*

The map $f:X\to \mathbb{R}\setminus \left\{0\right\}$ defined by $f\left(x,y\right)=x$ is a homeomorphism.

The function $f$
is the restriction of the projection, which is continuous. Hence,
$f$
is continuous. It is bijective with inverse given by
${f}^{-1}\left(x\right)=\left(x,1\u2215x\right)$, which is well-defined
and continuous for $x\ne 0$

If $I$ and
$J$ are two
subsets of $\mathbb{R}$ and
$I\times J$ is path-connected
then $I$
and $J$
are both intervals.
*Choice (a) is Correct!*
*Choice (b) is Incorrect.*

The statement is true. We may just consider the projection
$I\times J\to I$ and
$I\times J\to J$.
Those are surjective and continuous. It follows that the images
$I$ and
$J$ are path-connected, but
path-connected subset of $\mathbb{R}$
are exactly the intervals.

Consider the following subset $X$
in ${\mathbb{R}}^{2}$
How many $2$-points
does $X$
have?
*Choice (a) is Incorrect.*
*Choice (b) is Incorrect.*
*Choice (c) is Incorrect.*
*Choice (d) is Incorrect.*
*Choice (e) is Correct!*

Indeed, there are
infinitely may 2-points along the three line segments outside the triangle.

Consider the following subset $X$
in ${\mathbb{R}}^{2}$

From which of the following subsets can $X$ be distinguished via the cardinality of the set of $n$-points for a suitable $n\in \mathbb{N}$.
*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!*

From which of the following subsets can $X$ be distinguished via the cardinality of the set of $n$-points for a suitable $n\in \mathbb{N}$.

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be True.

For example, choice (d) should be False.

This subset is actually homeomorphic to
$X$ (a
homeomorphism is given by a horizontal reflection).

*True*This subset has no 3-point in contrast to $X$.*False*The number of $n$-points is the same as for $X$ ($\infty $, $1$, $0$, $0$, ... for $n=2,3,4,5,...$).*True*This subset has two 3-point in contrast to $X$, which has one.*False*This subset is actually homeomorphic to $X$ (a homeomorphism is given by a horizontal reflection).

Given two subset $X,Y\subset {\mathbb{R}}^{n}$.
Which of the following 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.
*Correct!*

For example, choice (a) should be True.

Choose a point $z\in X\cap Y$ and an
arbitrary point ${x}_{0}\in X\cup Y$ then
there exists a path ${\sigma}_{0}$
between ${x}_{0}$
and $z$ in
$X\cup Y$. Indeed, assume
${x}_{0}\in X$ then the path-connectedness
of $X$ implies the existence
of a path in $X$, which is also
a path in $X\cup Y$. Similar for
${x}_{0}\in Y$. In the same manner we
obtain a path ${\sigma}_{1}$ between
$z$ and any other point
${x}_{1}\in X\cup Y$. Now, the concatenation
of path gives a path ${\sigma}_{0}\ast {\sigma}_{1}$
from ${x}_{0}$
(via $z$)
to ${x}_{1}$.

For example, choice (b) should be False.

Consider, e.g. the upper and lower semi-circle. They are both path-connected, but the
intersection $\left\{\left(-1,0\right),\left(1,0\right)\right\}$
is not path-connnected.

*True*Choose a point $z\in X\cap Y$ and an arbitrary point ${x}_{0}\in X\cup Y$ then there exists a path ${\sigma}_{0}$ between ${x}_{0}$ and $z$ in $X\cup Y$. Indeed, assume ${x}_{0}\in X$ then the path-connectedness of $X$ implies the existence of a path in $X$, which is also a path in $X\cup Y$. Similar for ${x}_{0}\in Y$. In the same manner we obtain a path ${\sigma}_{1}$ between $z$ and any other point ${x}_{1}\in X\cup Y$. Now, the concatenation of path gives a path ${\sigma}_{0}\ast {\sigma}_{1}$ from ${x}_{0}$ (via $z$) to ${x}_{1}$.*False*Consider, e.g. the upper and lower semi-circle. They are both path-connected, but the intersection $\left\{\left(-1,0\right),\left(1,0\right)\right\}$ is not path-connnected.