Quotient spaces and equivalence relations

• Mar 8th 2009, 03:47 PM
Amanda1990
Quotient spaces and equivalence relations
Let I(x) denote the integer part of a real number x (ie the unique integer n such that $n \leq x < n+1$).

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

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

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

Consider the map $I: {R} \rightarrow {Z}$ by $x \rightarrow I(x)$. Is the map I continuous (when Z is endowed with the induced topology)?
• Mar 9th 2009, 04:19 AM
Opalg
Quote:

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

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

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

If $p(n) \in U$ then $n\in p^{-1}(U)$ which, by the definition of the quotient topology, is an open set in R and therefore contains $n-\delta$ (for $\delta$ small enough). Hence $n-1 = p(n-\delta) \in U$.

Quote:

Originally Posted by Amanda1990
Consider the map $I: {R} \rightarrow {Z}$ by $x \rightarrow I(x)$. Is the map I continuous (when Z is endowed with the induced topology)?

Notice that $1-\tfrac1n\to1$ as $n\to\infty$. But $p(1-\tfrac1n)=0$ and p(1) = 1.
• Mar 9th 2009, 05:19 AM
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 $p(x) = [I(x), I(x) +1)$?

On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from ${R}/$~ $\rightarrow Z$ (I think this i fairly clear). What is the topology on Z making J a homomorphism?
• Mar 9th 2009, 10:11 AM
Opalg
Quote:

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 $p(x) = [I(x), I(x) +1)$?

On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from ${R}/$~ $\rightarrow Z$ (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.
• Mar 9th 2009, 02:58 PM
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 (?)
• Mar 10th 2009, 02:01 AM
Opalg
Quote:

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 $\{k\in\mathbb{Z}:k\leqslant n\}$. I suppose I should have written it as $(-\infty,n]\cap\mathbb{Z}$.