# Thread: Quotient spaces and equivalence relations

1. ## Quotient spaces and equivalence relations

Let I(x) denote the integer part of a real number x (ie the unique integer n such that $\displaystyle n \leq x < n+1$).

I can show that $\displaystyle x$ ~ $\displaystyle y$ iff $\displaystyle I(x) = I(y)$ is an equivalence relation.

Let $\displaystyle p: {R} \rightarrow {R} /$ ~ be the quotient map, let $\displaystyle {R}/$~ be endowed with the quotient topology, and let U be an open set in $\displaystyle {R}/$~. Prove that if $\displaystyle n \in Z$ is such that $\displaystyle p(n) \in U$ then $\displaystyle p(n-1) \in U$.

Deduce that the open sets in $\displaystyle {R}/$~are the empty set, $\displaystyle {R}/$~ and the image sets $\displaystyle p(-\infty, n]$ where $\displaystyle n \in Z$.

Consider the map $\displaystyle I: {R} \rightarrow {Z}$ by $\displaystyle x \rightarrow I(x)$. Is the map I continuous (when Z is endowed with the induced topology)?

2. Originally Posted by Amanda1990
Let I(x) denote the integer part of a real number x (ie the unique integer n such that $\displaystyle n \leq x < n+1$).

I can show that $\displaystyle x$ ~ $\displaystyle y$ iff $\displaystyle I(x) = I(y)$ is an equivalence relation.

Let $\displaystyle p: {R} \rightarrow {R} /$ ~ be the quotient map, let $\displaystyle {R}/$~ be endowed with the quotient topology, and let U be an open set in $\displaystyle {R}/$~. Prove that if $\displaystyle n \in Z$ is such that $\displaystyle p(n) \in U$ then $\displaystyle p(n-1) \in U$.
If $\displaystyle p(n) \in U$ then $\displaystyle n\in p^{-1}(U)$ which, by the definition of the quotient topology, is an open set in R and therefore contains $\displaystyle n-\delta$ (for $\displaystyle \delta$ small enough). Hence $\displaystyle n-1 = p(n-\delta) \in U$.

Originally Posted by Amanda1990
Consider the map $\displaystyle I: {R} \rightarrow {Z}$ by $\displaystyle x \rightarrow I(x)$. Is the map I continuous (when Z is endowed with the induced topology)?
Notice that $\displaystyle 1-\tfrac1n\to1$ as $\displaystyle n\to\infty$. But $\displaystyle p(1-\tfrac1n)=0$ and p(1) = 1.

3. Thanks a lot - for the first part of the question I had the right idea but just couldn't write it properly.

For the last point you made, did you mean to use I instead of p? The map p just takes any real number to its equivalence class right? ie isn't $\displaystyle p(x) = [I(x), I(x) +1)$?

On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from $\displaystyle {R}/$~ $\displaystyle \rightarrow Z$ (I think this i fairly clear). What is the topology on Z making J a homomorphism?

4. Originally Posted by Amanda1990
Thanks a lot - for the first part of the question I had the right idea but just couldn't write it properly.

For the last point you made, did you mean to use I instead of p? The map p just takes any real number to its equivalence class right? ie isn't $\displaystyle p(x) = [I(x), I(x) +1)$?

On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from $\displaystyle {R}/$~ $\displaystyle \rightarrow Z$ (I think this i fairly clear). What is the topology on Z making J a homomorphism?
Yes, I should have said I rather than p.

The topology on Z making J a homeomorphism is given by the middle part of the question: The open sets are the intervals (–∞,n], together with the empty set and the whole space.

5. Well this is it - it looks as though the middle part of the question should give the answer to this but surely a topology on Z must be a family of subsets of Z (by definition). Then intervals (–∞,n] are not subsets of Z and so cannot form a topology (?)

6. Originally Posted by Amanda1990
Well this is it - it looks as though the middle part of the question should give the answer to this but surely a topology on Z must be a family of subsets of Z (by definition). Then intervals (–∞,n] are not subsets of Z and so cannot form a topology (?)
They certainly have to be subsets of Z, so you should interpret (–∞,n] as meaning $\displaystyle \{k\in\mathbb{Z}:k\leqslant n\}$. I suppose I should have written it as $\displaystyle (-\infty,n]\cap\mathbb{Z}$.