Quick question on basic topology
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Quick question on basic topology
Could someone oblige me and give a hint for a homeomorphism between [imath]\mathbb{Q}[/imath] and [imath]\mathbb{Q}\cap(0,\infty)[/imath]. Thank you.
On a side note, does anyone know if AC is required to prove that every secondcountable space is separable or can this be done without it?
On a side note, does anyone know if AC is required to prove that every secondcountable space is separable or can this be done without it?
 jestingrabbit
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Re: Quick question on basic topology
Harg wrote:Could someone oblige me and give a hint for a homeomorphism between [imath]\mathbb{Q}[/imath] and [imath]\mathbb{Q}\cap(0,\infty)[/imath].
Think rational functions with small powers of x, and maybe a little bit of piecewise too.
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Re: Quick question on basic topology
I've been thinking about it like that. What I've tried to do is to contract one ray of the rationals into an interval and then extend it on the other side with a linear function or something. The problem I keep running into is that I loose bijectivity on the nonlinear side.
 jestingrabbit
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Re: Quick question on basic topology
You might have more luck trying to construct it in the other direction and then taking the inverse.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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Re: Quick question on basic topology
There's probably some really clever way of doing it with a nice formula, but it's pretty easy to construct a piecewise linear bijection.
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 Yakk
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Re: Quick question on basic topology
What about the isomorphism that takes the group (R,+,0) to the group (R^{+},*,1)? That generates a topological isomorphism as well, I think, admittedly on a slightly different space.
And you can use it as inspiration to solve the problem for Q (admittedly, it won't be a group isomorphism I don't think!)
And you can use it as inspiration to solve the problem for Q (admittedly, it won't be a group isomorphism I don't think!)
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Quick question on basic topology
skeptical scientist wrote:There's probably some really clever way of doing it with a nice formula
How about = 1/(y  x)?
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Re: Quick question on basic topology
Yakk wrote:What about the isomorphism that takes the group (R,+,0) to the group (R^{+},*,1)? That generates a topological isomorphism as well, I think, admittedly on a slightly different space.
And you can use it as inspiration to solve the problem for Q (admittedly, it won't be a group isomorphism I don't think!)
The nice isomorphisms from (R,+,0) to the group (R^{+},*,1) are of the form [imath]x \mapsto a^x[/imath] for some positive a different from 1. Clearly that doesn't induce a homeomorphism from Q to Q^{+}. Moreover, (Q,+) and (Q^{+},*) are not isomorphic as groups, so the same trick wouldn't work. I'm not quite sure what you mean by using it as inspiration here.
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Re: Quick question on basic topology
ihope127 wrote:skeptical scientist wrote:There's probably some really clever way of doing it with a nice formula
How about = 1/(y  x)?
...well, if that isn't the most unusual equation I've ever seen. I meant y = 1/(y  x). I see now, though, that that isn't much of a bijection:
1/y = y  x
y  1/y = x
y^2  xy  1 = 0
y = (x + sqrt(x^2 + 4))/2
I'll have to try again.
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 Yakk
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Re: Quick question on basic topology
skeptical scientist wrote:Yakk wrote:What about the isomorphism that takes the group (R,+,0) to the group (R^{+},*,1)? That generates a topological isomorphism as well, I think, admittedly on a slightly different space.
And you can use it as inspiration to solve the problem for Q (admittedly, it won't be a group isomorphism I don't think!)
The nice isomorphisms from (R,+,0) to the group (R^{+},*,1) are of the form [imath]x \mapsto a^x[/imath] for some positive a different from 1. Clearly that doesn't induce a homeomorphism from Q to Q^{+}. Moreover, (Q,+) and (Q^{+},*) are not isomorphic as groups, so the same trick wouldn't work. I'm not quite sure what you mean by using it as inspiration here.
Under the group isomorphism mentioned, R maps to (0,1), {0} maps to {1}, and R+ maps to (1, infinity). That's about the only inspiration.
It isn't that hard to change those Rs to Qs:
Now, a: Q^{}>(0,1) in Q := x > 1/(x+1) maps Q^{} to (0,1) in Q topologically nicely.
And b: Q^{+}>(1, infinity) in Q := x > x+1 maps Q^{+} to (1, infinity) topologically nicely.
They do not, however, behave topologically nicely near 1. And that won't do.
The first step is obvious: fill in the gap.
And c: {0} > {1} defined in the only way.
The next step is to make the stitch work better. We need small negative numbers to map to a number near 1, instead of near 0. That's easy:
Take d: Q > (0,1) defined by x > 1a(x).
Then each of b, c, and d map a distinct portion of Q to a distinct portion of Q+, and the limits along the points that they stitch together behave topologically nicely.
Edit: I missed a +1 and a .
Last edited by Yakk on Sat Nov 15, 2008 4:40 pm UTC, edited 1 time in total.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Quick question on basic topology
Thanks to everyone. I had to go for a drive across half the country after jestingrabbit's second reply, so I had a lot of time to come up with [math]f(x)=\begin{cases}\frac{x1}{x};& 0<x\leq1\\x1;&x>1\end{cases}[/math]This seems to work and it looks so obvious now, but then it always does .
So, any ideas about the other thing?
So, any ideas about the other thing?
Re: Quick question on basic topology
Well, countable choice suffices to prove that every secondcountable space is separable. The converse is, as far as I can tell, not easily attacked. However, it's possible to prove [ref] that if every subspace of the reals (each such is certainly secondcountable) is separable, then countable choice holds for subsets of the reals. So you're clearly going to need some form of choice.
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Re: Quick question on basic topology
Ooh, strange paper. Shiny! *drools slightly*
Re: Quick question on basic topology
I would like to confirm this. Yes or no question:
is the topology generated by the subbasis [math]\mathcal{P}=\{[a,\infty)\subseteq\mathbb{R}\mid a\in\mathbb{R}\}\cup\{(\infty,b]\subseteq\mathbb{R}\mid b\in\mathbb{R}\}[/math] in fact the discrete topology?
is the topology generated by the subbasis [math]\mathcal{P}=\{[a,\infty)\subseteq\mathbb{R}\mid a\in\mathbb{R}\}\cup\{(\infty,b]\subseteq\mathbb{R}\mid b\in\mathbb{R}\}[/math] in fact the discrete topology?
 Yakk
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Re: Quick question on basic topology
Let x be an arbitrary element of R. Is {x} an open set?
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Quick question on basic topology
Ah, very nice. I didn't notice this characterization of the discrete topology before. I kinda just saw that this was strictly stronger than the Sörgenfrey topology on R and thought: "Well, Sörgenfrey is pretty big, so I suppose this could be discrete." Yes, thank you.
It is interesting, though, that adding a bit of points to the subbasis elements should give something so different from the euclidean topology.
It is interesting, though, that adding a bit of points to the subbasis elements should give something so different from the euclidean topology.

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Re: Quick question on basic topology
Even quicker, even more basic: the closure of the set of rationals is the set of reals, and the irrationals are boundary points of the set of rationals. True?
Re: Quick question on basic topology
Suffusion of Yellow wrote:The closure of the set of rationals is the set of reals
Yes: Q is dense in R.
Suffusion of Yellow wrote:The irrationals are boundary points of the set of rationals
Yes  in fact, the boundary of Q is the whole of R.
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Re: Quick question on basic topology
Token wrote:Suffusion of Yellow wrote:The closure of the set of rationals is the set of reals
Yes: Q is dense in R.Suffusion of Yellow wrote:The irrationals are boundary points of the set of rationals
Yes  in fact, the boundary of Q is the whole of R.
Cool, thanks  my school offers "Topology is a prerequisite" courses but curiously not a Topology course.
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