❖
✖ 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

Which of the following assertions are 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.
*There is at least one mistake.*

For example, choice (e) should be True.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be True.

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

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

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

For example, choice (a) should be True.

For example, choice (b) should be True.

For example, choice (c) should be True.

Take
an open cover $\mathcal{\mathcal{F}}$
of $X\cup Y$. Then
$\mathcal{\mathcal{F}}$ is also an open
cover for $X$ and
$Y$, respectively.
Compactness of $X$ and
$Y$ implies the existence
of finite subcovers ${\mathcal{\mathcal{F}}}_{X}$
and ${\mathcal{\mathcal{F}}}_{Y}$ of
$X$ and
$Y$, respectively.
Then ${\mathcal{\mathcal{F}}}_{X}\cup {\mathcal{\mathcal{F}}}_{Y}\subset \mathcal{\mathcal{F}}$ is a finite
subcover for $X\cup Y$.

For example, choice (d) should be False.

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.

For example, choice (f) should be False.

Consider $\left[0,1\right]=\left(-\infty ,1\right]\cap \left[0,\infty \right)$.

*True*This is one direction of Theorem 5.11.*True*This is the other direction of Theorem 5.11.*True*Take an open cover $\mathcal{\mathcal{F}}$ of $X\cup Y$. Then $\mathcal{\mathcal{F}}$ is also an open cover for $X$ and $Y$, respectively. Compactness of $X$ and $Y$ implies the existence of finite subcovers ${\mathcal{\mathcal{F}}}_{X}$ and ${\mathcal{\mathcal{F}}}_{Y}$ of $X$ and $Y$, respectively. Then ${\mathcal{\mathcal{F}}}_{X}\cup {\mathcal{\mathcal{F}}}_{Y}\subset \mathcal{\mathcal{F}}$ is a finite subcover for $X\cup Y$.*False*Consider $\left[0,1\right]=\left[0,1\right)\cup \left(0,1\right]$.*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 $\left[0,1\right]=\left(-\infty ,1\right]\cap \left[0,\infty \right)$.

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.
*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 True.
*Correct!*

For example, choice (a) should be True.

For example, choice (b) should be True.

For example, choice (c) should be False.

For example, choice (d) should be True.

A closed subset
$A\subset X$ is compact, since
$X$ is compact. Then its
image $f\left(A\right)$ is compact,
since $f$ is continuous.
Now, it follows that $f\left(A\right)$
is closed, since $Y$
is Hausdorff.

*True*This is the definition of continuity of $f$.*True*This is equivalent to the continuity of $f$.*False*Take $f:\left\{0\right\}\to \mathbb{R}$ the inclusion map. Then $f\left(\left\{0\right\}\right)=\left\{0\right\}$, which is not open in $\mathbb{R}$.*True*A closed subset $A\subset X$ is compact, since $X$ is compact. Then its image $f\left(A\right)$ is compact, since $f$ is continuous. Now, it follows that $f\left(A\right)$ 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!*
*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 $\tau $
and ${\tau}^{\prime}$ on the same
topological space $X$ such that
${\tau}^{\prime}\u228a\tau $. Which of the following
assertions 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.

For example, choice (a) should be True.

For example, choice (b) should be False.

Pick any $U\in \tau \setminus {\tau}^{\prime}$.
Then ${id}_{X}^{-1}\left(U\right)=U\notin {\tau}^{\prime}$.

Consider two topologies $\tau $
and ${\tau}^{\prime}$ on the same
topological space $X$ such that
${\tau}^{\prime}\subset \tau $. Which of the following
assertions are true?
*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.
*There is at least one mistake.*

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

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

For example, choice (a) should be False.

For example, choice (b) should be True.

For example, choice (c) should be True.

For example, choice (d) should be False.

For example, choice (e) should be False.

For example, choice (f) should be True.

Note that ${id}_{X}:\left(X,\tau \right)\to \left(X,{\tau}^{\prime}\right)$ is a
continuous bijection from a compact space to a Hausdorff space. Hence, it is a homeomorphism.
Therefore $U\in \tau \iff U={id}_{X}\left(U\right)\in {\tau}^{\prime}$.

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