An exercise on the quotient mapping $f:\mathbb{R}\to\mathbb{Z}$

Question

Let $f:\mathbb{R}\to\mathbb{Z}$ be a function such that each $x\in\mathbb{R}$ it associates the greatest integer less or equal to $x$. Suppose that $\mathbb{R}$ is endowed with the eucledian topology such that $\tau$ is the quotient topology in $\mathbb{Z}$ defined by $f$.

a) Compute $f^{-1}(n)$ for $n\in\mathbb{Z}$

b)Prove $\tau=\{\emptyset,\mathbb{Z}\}\cup\{]-\infty,n]\cap\mathbb{Z}:n\in\mathbb{Z}\}$

c) Determine if $(\mathbb{Z},\tau)$ is metrizable.

d)Determine if $(\mathbb{Z},\tau)$ is compact.

a) If $f^{-1}(n)=[n,n+1)$ once $\leqslant x\leqslant n+1$

b) If we have the infinite set (...,-n-1,...-3,-2,-1,0,1,2,3...n-1...)

Let $(\mathbb{R},\tau')$ be the topological space where $\tau'$ is the Euclidean topology

$f^{-1}(\mathbb{Z})=\mathbb{R}\in \tau'\implies\mathbb{Z}\in\tau $

$f^{-1}(\emptyset)=\emptyset\in \tau'\implies\emptyset\in\tau $

If I pick up a set in $\mathbb{Z}$ like }$(...,-n,...-3,-2,-1,0,1,2,3...n)$ then I prove it is open. $f^{-1}((...,-n,...-3,-2,-1,0,1,2,3...n))=...\cup [-n-1,-n)\cup[-3,-2)\cup[-2,-1)\cup[-1,0)\cup[1,2)\cup[3,4)...\cup[n-1,n)\cup=(-\infty,n)\in \tau'\implies (\infty,n)\cap\mathbb{Z}\in\tau$

Note the compliment of $\mathbb{Z}\setminus((\infty,n)\cap\mathbb{Z})=[n,\infty)\cap\mathbb{Z}$ which is closed hence I do need to try other sets.

This proves $\tau=\{\emptyset,\mathbb{Z}\}\cup\{]-\infty,n]\cap\mathbb{Z}:n\in\mathbb{Z}\}$ is the topology $\tau$.

c)

Proposition: Every metric space is Hausdorff.

Let $x,y\in\mathbb{Z}$ there is an open set $x\in U$ and $y\in V$ such that $U=(-\infty,j]\cap\mathbb{Z}$ and $V=(-\infty,j']\cap\mathbb{Z}$. It is trivial to notice $U\cap V\neq \emptyset$. Then as $j$ and $j'$ are arbitrary elements of $\mathbb{Z}$ it can be concluded that $\mathbb{Z},\tau$ is not Hausdorff hence it is not metrizable.

d)

Proposition: Every compact space is limited.

Since $\mathbb{Z},\tau$ is not limited hence it is not compact.

Question:

Is this answer right?

Answer 1

Indeed $f^{-1}[\{n\}]= [n, n+1)$. Note that this means that $\{n\}$ is not open in $\tau$, as the quotient topology means that $O \subseteq \mathbb{Z}$ is open if and only if $f^{-1}[O]$ is open in $\mathbb{R}$. And $[n,n+1)$ is not open.

Suppose $O$ is open in $\mathbb{Z}$ under $\tau$. This means $$f^{-1}[O]=\bigcup \{f^{-1}[\{n\}]: n \in O\} = \bigcup_{n \in O} [n,n+1)$$

is open in the Euclidean topology. If the integer $m$ is in $O$, so must $m-1$ be, as otherwise $m$ is not an interior point of $f^{-1}[O]$.

The only subsets $O$ of $\mathbb{Z}$ that obey $\forall M \in O: m-1 \in O$ are indeed $\emptyset, \mathbb{Z}$ and all sets of the form $(-\infty,n]$ for $n \in \mathbb{Z}$ and these sets indeed have open pre-images, so these form exactly the quotient topolgoy $\tau$.

As all non-empty open subsets of $\tau$ must intersect (clear from the form of the open sets, or note that $n \in O_1, m \in O_2$ implies $\min(n,m) \in O_1 \cap O_2$) $(\mathbb{Z},\tau)$ is not Hausdorff, so that idea checks out.

The open cover $(-\infty,1], (-\infty,2], \ldots$ has no finite subcover, so $(\mathbb{Z}, \tau)$ is not compact.

Answer 2

a) Correct. Though I don't know what you mean by "once $n\leq x\leq n+1$".

b) You have proved that $\tau$ contains the given topology, i.e. that it is at least as fine as the given topology. You have not shown equality; there could be other sets in $\tau$. I would suggest that you prove the implication $$n\in U\quad\Rightarrow\quad n-1\in U,$$ for all open sets $U\subset\Bbb{Z}$.

c) Correct, though poorly worded. I would suggest a proof by contradiction; suppose $(\Bbb{Z},\tau)$ is Hausdorff...

d) This is wrong, the proposition as stated is false, or at best very unclear. Perhaps you had in mind

Proposition: Every compact metric space is bounded.

Unfortunately you have just shown in part c that $(\Bbb{Z},\tau)$ is not a metric space, so the proposition does not apply. But it is not hard to give an explicit open cover that does not have a finite subcover; any infinite open cover that doesn't contain $\Bbb{Z}\in\tau$ will do, in fact.