❖
✖ 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 topologies on $\mathbb{R}$
are Hausdorff?
*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 False.
*There is at least one mistake.*

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

For example, choice (a) should be False.

Any two non-empty open subsets have the form
${U}_{1}=\mathbb{R}\setminus {F}_{1}$ and
${U}_{2}=\mathbb{R}\setminus {F}_{2}$ with
${F}_{1},{F}_{2}$ being finite.
Then ${U}_{1}\cap {U}_{2}=\mathbb{R}\setminus \left({F}_{1}\cup {F}_{2}\right)$. But
$\left({F}_{1}\cup {F}_{2}\right)$ is finite and
$\mathbb{R}$ is infinite. Hence,
${U}_{1}\cap {U}_{2}\ne \varnothing $.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be False.

The only open
subset containing $0$
is $U=\mathbb{R}$. Now, for every
open neighbourhood $V$
of another point $x\in \mathbb{R}$
we have $U\cap V=V\ne \varnothing $.

*False*Any two non-empty open subsets have the form ${U}_{1}=\mathbb{R}\setminus {F}_{1}$ and ${U}_{2}=\mathbb{R}\setminus {F}_{2}$ with ${F}_{1},{F}_{2}$ being finite. Then ${U}_{1}\cap {U}_{2}=\mathbb{R}\setminus \left({F}_{1}\cup {F}_{2}\right)$. But $\left({F}_{1}\cup {F}_{2}\right)$ is finite and $\mathbb{R}$ is infinite. Hence, ${U}_{1}\cap {U}_{2}\ne \varnothing $.*False*Again any two non-empty open subsets have the form ${U}_{1}=\left({a}_{1},\infty \right)$ and ${U}_{2}=\left({a}_{2},\infty \right)$ (where ${a}_{1},{a}_{2}\in \mathbb{R}\cup \left\{-\infty \right\}$). Then ${U}_{1}\cap {U}_{2}=\left(inf\left\{{a}_{1},{a}_{2}\right\},\infty \right)\ne emptyset$*False*Obviously, we have $0\in {U}_{1}\cap {U}_{2}$ for any two non-empty open subsets.*False*The only open subset containing $0$ is $U=\mathbb{R}$. Now, for every open neighbourhood $V$ of another point $x\in \mathbb{R}$ we have $U\cap V=V\ne \varnothing $.

Consder a continuous map $f:X\to Y$.
Which of the following statements are true? Think carefully about a justification for your
answer.
*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 (d) should be False.
*Correct!*

For example, choice (a) should be False.

For example, choice (b) should be True.

Indeed, for ${x}_{1},{x}_{2}\in X$ we have
$f\left({x}_{1}\right)\ne f\left({x}_{2}\right)$. Hence, there are
disjoint open subsets ${V}_{1},{V}_{2}\subset Y$
containing $f\left({x}_{1}\right)$ and
$f\left({x}_{2}\right)$, respectively.
Then ${f}^{-1}\left({V}_{1}\right)$ and
${f}^{-1}\left({V}_{2}\right)$ are open since
$f$ is continuous,
disjoint and contain ${x}_{1}$
and ${x}_{2}$, respectively.

For example, choice (d) should be False.

Take $f={id}_{\mathbb{R}}:\mathbb{R}\to \mathbb{R}$,
with the usual topology on the domain and the indiscrete topology on the codomain.

*False*Remark 4.9 gives a counterexample.*True*Indeed, for ${x}_{1},{x}_{2}\in X$ we have $f\left({x}_{1}\right)\ne f\left({x}_{2}\right)$. Hence, there are disjoint open subsets ${V}_{1},{V}_{2}\subset Y$ containing $f\left({x}_{1}\right)$ and $f\left({x}_{2}\right)$, respectively. Then ${f}^{-1}\left({V}_{1}\right)$ and ${f}^{-1}\left({V}_{2}\right)$ are open since $f$ is continuous, disjoint and contain ${x}_{1}$ and ${x}_{2}$, respectively.*True*as for injectivity.*False*Take $f={id}_{\mathbb{R}}:\mathbb{R}\to \mathbb{R}$, with the usual topology on the domain and the indiscrete topology on the codomain.

Which of the following quotient spaces are Hausdorff?
*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.
*Correct!*

For example, choice (a) should be False.

Note,
that $A=\left(-1,1\right)\in \mathbb{R}\u2215\left(-1,1\right)$
is a single point in the identification space. The preimage
${q}^{-1}\left(\left\{A\right\}\right)=\left(-1,1\right)$ is not closed
in $\mathbb{R}$. Hence,
$\left\{A\right\}$ is not
closed in $\mathbb{R}\u2215\left(-1,1\right)$,
but by Proposition 4.6 singletons are closed in Hausdorff spaces. By the same argument
$X\u2215A$ is never
Hausdorff if $A\subset X$
is not closed.

For example, choice (b) should be True.

Consider the continuous surjection $f:\mathbb{R}\to \mathbb{R}$
given by

$$f\left(t\right)=\left\{\begin{array}{cc}t-1\phantom{\rule{1em}{0ex}}\hfill & t\ge 1,\hfill \\ t+1\phantom{\rule{1em}{0ex}}\hfill & t\le -1,\hfill \\ 0\phantom{\rule{1em}{0ex}}\hfill & t\in \left[-1,1\right].\hfill \end{array}\right.$$

This induces a continuous bijection (which is actually even a homeomorphism)
$\mathbb{R}\u2215\left[-1,1\right]\to \mathbb{R}$. Hence, by the
previous question $\mathbb{R}\u2215\left[-1,1\right]$
is Hausdorff.
For example, choice (c) should be True.

Consider
$X=\left\{\left(x,y\right)\mid y=1\text{or}{x}^{2}+{y}^{2}=1\right\}\subset {\mathbb{R}}^{2}$. Then the continuous
surjection $f:\mathbb{R}\to X$,

$$f\left(t\right)=\left\{\begin{array}{cc}\left(t-1,1\right)\phantom{\rule{1em}{0ex}}\hfill & t\ge 1,\hfill \\ \left(t+1,1\right)\phantom{\rule{1em}{0ex}}\hfill & t\le -1,\hfill \\ \left(sin\left(2\pi t\right),cos\left(2\pi t\right)\right)\phantom{\rule{1em}{0ex}}\hfill & t\in \left[-1,1\right].\hfill \end{array}\right.$$

Induces a continuous bijection $\mathbb{R}\u2215\left\{-1,1\right\}\to X$
(which is actually a homeomorphism). Then the Hausdorff property follows by the
previous question. *False*Note, that $A=\left(-1,1\right)\in \mathbb{R}\u2215\left(-1,1\right)$ is a single point in the identification space. The preimage ${q}^{-1}\left(\left\{A\right\}\right)=\left(-1,1\right)$ is not closed in $\mathbb{R}$. Hence, $\left\{A\right\}$ is not closed in $\mathbb{R}\u2215\left(-1,1\right)$, but by Proposition 4.6 singletons are closed in Hausdorff spaces. By the same argument $X\u2215A$ is never Hausdorff if $A\subset X$ is not closed.*True*Consider the continuous surjection $f:\mathbb{R}\to \mathbb{R}$ given by$$f\left(t\right)=\left\{\begin{array}{cc}t-1\phantom{\rule{1em}{0ex}}\hfill & t\ge 1,\hfill \\ t+1\phantom{\rule{1em}{0ex}}\hfill & t\le -1,\hfill \\ 0\phantom{\rule{1em}{0ex}}\hfill & t\in \left[-1,1\right].\hfill \end{array}\right.$$This induces a continuous bijection (which is actually even a homeomorphism) $\mathbb{R}\u2215\left[-1,1\right]\to \mathbb{R}$. Hence, by the previous question $\mathbb{R}\u2215\left[-1,1\right]$ is Hausdorff.

*True*Consider $X=\left\{\left(x,y\right)\mid y=1\text{or}{x}^{2}+{y}^{2}=1\right\}\subset {\mathbb{R}}^{2}$. Then the continuous surjection $f:\mathbb{R}\to X$,$$f\left(t\right)=\left\{\begin{array}{cc}\left(t-1,1\right)\phantom{\rule{1em}{0ex}}\hfill & t\ge 1,\hfill \\ \left(t+1,1\right)\phantom{\rule{1em}{0ex}}\hfill & t\le -1,\hfill \\ \left(sin\left(2\pi t\right),cos\left(2\pi t\right)\right)\phantom{\rule{1em}{0ex}}\hfill & t\in \left[-1,1\right].\hfill \end{array}\right.$$Induces a continuous bijection $\mathbb{R}\u2215\left\{-1,1\right\}\to X$ (which is actually a homeomorphism). Then the Hausdorff property follows by the previous question.

For which of the following equivalence relations is the quotient space
*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 False.
*There is at least one mistake.*

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

$$\left(\mathbb{R}\times \left\{-1,1\right\}\right)\u2215\sim $$

Hausdorff?
For example, choice (a) should be True.

For example, choice (b) should be True.

Consider $X=\left\{\left(x,y\right)\mid x=-y\text{or}x=y\right\}\subset {\mathbb{R}}^{2}$.
The continuous surjection

$$f:\mathbb{R}\times \left\{-1,1\right\}\to X,\phantom{\rule{2.77695pt}{0ex}}\left(x,a\right)\mapsto \left(x,ax\right).$$

induces a continuous bijection $\left(\mathbb{R}\times \left\{-1,1\right\}\right)\u2215\sim \to X$
(which is actually a homeomorphism). By Question 2 the Hausdorff property follows.
For example, choice (c) should be False.

For example, choice (d) should be False.

For example, choice (e) should be True.

Consider

$$X=\left\{-1\right\}\times \mathbb{R}\cup \left\{1\right\}\times \mathbb{R}\cup \left[-1,1\right]\times \left\{0\right\}\subset {\mathbb{R}}^{2}.$$

The continuous surjection $f:\mathbb{R}\times \left\{-1,1\right\}\to X$
given by
$$f\left(x,a\right)=\left\{\begin{array}{cc}\left(1,a\left(x-1\right)\right)\phantom{\rule{1em}{0ex}}\hfill & x\ge 1,\hfill \\ \left(-1,a\left(x-1\right)\right)\phantom{\rule{1em}{0ex}}\hfill & x\le -1,\hfill \\ \left(x,0\right)\phantom{\rule{1em}{0ex}}\hfill & x\in \left[-1,1\right].\hfill \end{array}\right.$$

induces a continuous bijection $\left(\mathbb{R}\times \left\{-1,1\right\}\right)\u2215\sim \to X$
(which is actually a homeomorphism). Again, the Hausdorff property follows from
Question 2. *True*This space can be easily seen to be homeomorphic to $\mathbb{R}$. Hence, it is Hausdorff. The homeomorphism is induced by the projection to the first factor.*True*Consider $X=\left\{\left(x,y\right)\mid x=-y\text{or}x=y\right\}\subset {\mathbb{R}}^{2}$. The continuous surjection$$f:\mathbb{R}\times \left\{-1,1\right\}\to X,\phantom{\rule{2.77695pt}{0ex}}\left(x,a\right)\mapsto \left(x,ax\right).$$induces a continuous bijection $\left(\mathbb{R}\times \left\{-1,1\right\}\right)\u2215\sim \to X$ (which is actually a homeomorphism). By Question 2 the Hausdorff property follows.

*False*The $q\left(\left(0,1\right)\right)$ and $q\left(\left(0,-1\right)\right)$ cannot be separated by open subset. The proof is very similar to Remark 4.9 in the lecture notes.*False*The equivalence classes $\left[\left(1,1\right)\right]$ and $\left[\left(1,-1\right)\right]$ (here $\left(1,1\right)$ and $\left(1,-1\right)$ denote the pairs not the open intervals) can not be separated by open subsets. The proof is again similar to Remark 4.9 in the lecture notes.*True*Consider$$X=\left\{-1\right\}\times \mathbb{R}\cup \left\{1\right\}\times \mathbb{R}\cup \left[-1,1\right]\times \left\{0\right\}\subset {\mathbb{R}}^{2}.$$The continuous surjection $f:\mathbb{R}\times \left\{-1,1\right\}\to X$ given by$$f\left(x,a\right)=\left\{\begin{array}{cc}\left(1,a\left(x-1\right)\right)\phantom{\rule{1em}{0ex}}\hfill & x\ge 1,\hfill \\ \left(-1,a\left(x-1\right)\right)\phantom{\rule{1em}{0ex}}\hfill & x\le -1,\hfill \\ \left(x,0\right)\phantom{\rule{1em}{0ex}}\hfill & x\in \left[-1,1\right].\hfill \end{array}\right.$$induces a continuous bijection $\left(\mathbb{R}\times \left\{-1,1\right\}\right)\u2215\sim \to X$ (which is actually a homeomorphism). Again, the Hausdorff property follows from Question 2.