Conversation started Feb 13, 2023 at 17:44.
jrh
jrh
Feb 13, 2023 17:44
I've been working through "A Guide to Proof Writing", on page 497 of the proof in Example 3, the author assumes that b is in the intersect of X and Z but I am not sure how that's a safe assumption, couldn't Z be the empty set? That would make the proposition true but it would be a false assumption.
Ted Shifrin
Ted Shifrin
You don't need to do this.
Define the map by sending $[x_1,x_2,x_3]$ to $[0,x_1,x_2,x_3]$. Then well-definedness is trivial.
jrh
jrh
The example is trying to prove "for all sets X, Y, and Z, if X is a subset of Y, then (X intersect Z) is a subset of (Y intersect Z)"
Koro
Koro
@TedShifrin: is the idea correct? Once we show these two maps to be inverses of each other, we have shown the homeomorphism and we are done :-).
leslie townes
leslie townes
jrh: Z could be empty, but in that case there is nothing to prove. the implication "if something is in X intersect Z, then ___" is true no matter what ___ is in that case.
Sine of the Time
Sine of the Time
@TedShifrin Ok, but this is still considered inclusion map? because to proof it's a deformation retract I've to proof $i\circ r\simeq Id$
Ted Shifrin
Ted Shifrin
Feb 13, 2023 17:45
The inclusion and the retraction are never inverses.
Koro
Koro
@TedShifrin that’s our i.
Ted Shifrin
Ted Shifrin
They are homotopy inverses, but not set inverses.
Koro
Koro
Ohh, but sine said that they want to show $P^3-p\simeq P^2$. I suppose that $\simeq$ was for homeomorphism, was it not?
Sine of the Time
Sine of the Time
@Koro homotopy, the're not homeomorphic I think
Ted Shifrin
Ted Shifrin
Absolutely not homeomorphic. One is $2$ complex-dimensional, the other is $3$.
Koro
Koro
Feb 13, 2023 17:48
Ahh okay.
leslie townes
leslie townes
jrh: more generally to prove that a set A is a subset of set B, it's enough to prove for all a that if a is in A, then a is in B. A being empty doesn't 'break' this reasoning, A being empty is just the one case in which element-wise reasoning would not be needed.
jrh
jrh
That is fair, the empty set is a subset of any set. Another thing I don't completely understand about the author's method, it seems like the procedure assumes q in p -> q; though if p were false, wouldn't that give false positive proof results?
Novice
Novice
@onepotatotwopotato Not an important question, but what were your analysis and manifolds chops like before looking at Jost's RG book?
Sine of the Time
Sine of the Time
@TedShifrin do you have 5 minutes to check if my reasoning is correct?
Ted Shifrin
Ted Shifrin
If it's correct, it should only take a few seconds, not 5 minutes :)
Sine of the Time
Sine of the Time
Feb 13, 2023 17:55
@TedShifrin yes, just give me the time to write it :(
Ted Shifrin
Ted Shifrin
Are you starting with the copy of $\Bbb P^2$ I suggested?
leslie townes
leslie townes
jrh: if p is false, then p -> q is true. this is slightly different from "assuming" q. it's only noting that q is implied by something false.
Ted Shifrin
Ted Shifrin
If Leslie always tells the truth, then the moon is green.
Sine of the Time
Sine of the Time
I want to prove $X \simeq Y$, where $X=\Bbb{P}^2$ and $Y=\Bbb{P}^3\setminus \{[1,0,0,0]\}$. Consider $i: X\to Y$ defined letting $i([x_0,x_1,x_2])=[0,x_0,x_1,x_2]$ and $r:Y \to X$ defined letting $r([x_0,x_1,x_2,x_3])=[x_0,x_1,x_2]$. Is it true that $i \circ r\simeq Id_Y$? If we consider $F:Y\times I \to Y$ defined by $F([x_0,x_1,x_2,x_3],t)=[tx_0,x_1,x_2,x_3]$ we should have the homotopy
Ted Shifrin
Ted Shifrin
NO, your map is wrong.
Sine of the Time
Sine of the Time
Feb 13, 2023 18:01
@TedShifrin are you referring to $F$?
Ted Shifrin
Ted Shifrin
You need to project from the point $[1,0,0,0]$ to the plane $x_0=0$.
I'm referring to $r$. You need to draw a picture.
Sine of the Time
Sine of the Time
but $[1,0,0,0]$ is not in the domain
jrh
jrh
Right, if p were false that'd be a trivial proof. So we're assuming X is a subset of Y, and then assuming b is in (X intersect Z), the only thing left to prove is that b is in (Y intersect Z). I guess that makes sense, when you let the trivial cases of subset and implication fall away. Thanks.
Ted Shifrin
Ted Shifrin
That's the point. The picture of the mapping is that you're projecting from $p$. You send $q$ to the point where the line $pq$ intersects the image plane.
 
Conversation ended Feb 13, 2023 at 18:02.