sequences and projections

[RichD]

Given an infinite sequence {P[n]}, in the x-y plane,
which converges to a limit point L. Project those points
onto the x-axis, i.e. their x coordinates, creating the
sequence {x[n]}. Finally, project L onto the x axis,
call this point M.

Consider:
i) M is the limit of {x[n]}
ii) The limit of {x[n]} is M.

Do both these assertions qualify as theorems?

--
Rich

[Barry Schwarz]

What is the difference between the two?

Is it a theorem if it follows directly from the definition?

--
Remove del for email

[FromTheRafters]

Barry Schwarz explained on 1/21/2022 :

> On Thu, 20 Jan 2022 19:53:42 -0800 (PST), RichD
> <r_dela...@yahoo.com> wrote:
>
>> Given an infinite sequence {P[n]}, in the x-y plane,
>> which converges to a limit point L. Project those points
>> onto the x-axis, i.e. their x coordinates, creating the
>> sequence {x[n]}. Finally, project L onto the x axis,
>> call this point M.
>>
>> Consider:
>> i) M is the limit of {x[n]}
>> ii) The limit of {x[n]} is M.
>>
>> Do both these assertions qualify as theorems?
>
> What is the difference between the two?

As stated, they seem the same to me. The most wonderful thing about
Tiggers is that Tiggers are wonderful things.

> Is it a theorem if it follows directly from the definition?

I'm unfamiliar with this area, is such a projection a subsequence which
necessarily also converges to L?

[Mike Terry]

It is a theorem that if P[n] converges, then the projections converge to the projection of the limit
of P[n]. It's a consequence of how the metric on R^2 relates to the metric on R.

The reverse also holds: if x[n] and y[n] converge to X and Y respectively, then (x[n], y[n])
converges (according to the R^2 metric) to (X,Y).

Both of these results require proof - they are not correct simply by definition, but of course they
follow (directly?) from the definitions being used, as is the case for all theorems!

Mike.

[sergio]

not theorems, just a problem.

M is simply L(M,K), where K is the y limit point

x and y are orthogonal, so you can separate them into 2 sequences

[RichD]

On January 20, RichD wrote:
> Given an infinite sequence {P[n]}, in the x-y plane,
> which converges to a limit point L. Project those points
> onto the x-axis, i.e. their x coordinates, creating the
> sequence {x[n]}. Finally, project L onto the x axis,
> call this point M.
>
> Consider:
> i) M is the limit of {x[n]}
> ii) The limit of {x[n]} is M.

hmmm.. trying to present a tricky problem, I flubbed the phrasing.
It's defective.

Re-try:
Given an infinite sequence {P[n]}, in the x-y plane. Project those points

onto the x-axis, i.e. their x coordinates, creating the sequence {x[n]}

Now:
i) The limit of {x[n]} is also the projection of the limit of {P[n]}
ii) The x-projection of the limit of {P[n]} is also the limit of {x[n]}

True/false/maybe? Counterexamples?


--
Rich

[Mike Terry]

They both still say the same thing, because saying A=B is always equivalent to saying B=A.

You've dropped the requirement that P[n] converges, which changes things a bit. Let's call the x-
and y-projection functions f_x and f_y respectively...

IF P[n] CONVERGES (in the plane) then BOTH f_x(P[n]) and f_y(P[n]) will converge, and

lim [ f_x( P[n] ) ] = f_x( lim [ P[n] ] ) [A]
lim [ f_y( P[n] ) ] = f_y( lim [ P[n] ] ) [B]

Conversely, if BOTH x[n] and y[n] converge, then P[n] = (x[n], y[n]) will converge, and

lim [ P[n] ] = lim [ (x[n],y[n]) =
= ( lim [x[n]], lim [y[n] ) [C]

Those results can be proved from the definition of convergence in the plane R^2, and convergence in
R. (If you're familiar with metric spaces, then there is just one definition of "convergence" for
metric spaces, but in your scenario we have two metric spaces involved, R and R^2, each with their
own metric defined.)

If a pre-condition above does not hold, the conclusion will not necessarily hold - e.g. if P[n] does
NOT converge, it would still be possible for f_x(P[n]) to converge, but if it does then we could be
sure f_y(P[n]) does not converge, and vice-versa swapping x,y. E.g. when P[n] = ( 0, n ) obviously
f_x( P[n] ) = 0, which converges to 0, but f_x( P[n] ) = n which does not converge...
(alternatively, maybe /neither/ projected sequence converges, as when P[n] = ( n, n ).)

The projection functions f_x and f_y are /continuous/ functions between R^2 and R, and the above
results are consequences of this.

Hmm, perhaps there is confusion on the use of the term "the limit of..."? We don't use that term,
or the expression "lim [ P[n] ]" unless we know that the sequence involved converges... (That's why
I was careful in my wording for the results above, e.g. not writing [A] until after I've stated that
both P[n] and f_x(P[n]) converge. An expression like f_x( lim [ P[n] ] ) is meaningless when P[n]
does not converge, and so shouldn't even be uttered! :) )


Mike.

[Chris M. Thomasson]

For some damn reason, this is making me thing of the following simple
setup. Very quick, and I must be missing something important, anyway:

My Test Code
________________________
namespace ct_project_test // just for fun
{

void
limit_line(
ct::plot::cairo::plot_2d& plot,
glm::vec2 p0_xy_plane,
glm::vec2 p1_xy_plane,
glm::vec2 p0_x_plane,
glm::vec2 p1_x_plane,
unsigned long n
) {
glm::vec2 dif_xy_plane = p1_xy_plane - p0_xy_plane;
glm::vec2 dif_x_plane = p1_x_plane - p0_x_plane;

float normal_base = 1.f / (n - 1);

plot.line(p0_xy_plane, p1_xy_plane, CT_RGBF(1, 1, 0));
plot.line(p0_x_plane, p1_x_plane, CT_RGBF(1, 0, 0));

for (unsigned long i = 0; i < n; ++i)
{
float normal = normal_base * i;

glm::vec2 p0_xy = p0_xy_plane + dif_xy_plane * normal;
glm::vec2 p0_x = p0_x_plane + dif_x_plane * normal;

plot.circle_filled(p0_xy, .03f, CT_RGBF(1, 0, 0));
plot.circle_filled(p0_x, .03f, CT_RGBF(1, 1, 0));

plot.line(p0_xy, p0_x, CT_RGBF(1, 0, 1));

std::cout << "p0_xy = (" << p0_xy.x << ", " << p0_xy.y <<
") projected to ";
std::cout << "p0_x = (" << p0_x.x << ", " << p0_x.y << ")\n";
}
}


void manifest(ct::plot::cairo::plot_2d& plot)
{
std::cout << "ct_project_test::manifest\n\n";

{
glm::vec2 l0_p0 = { -1, 0 };
glm::vec2 l0_p1 = { -1, 1 };

glm::vec2 l1_p0 = { -1, 0 };
glm::vec2 l1_p1 = { 1, 0 };

unsigned long n = 10;

limit_line(plot, l0_p0, l0_p1, l1_p0, l1_p1, n);
}
}
}
________________________


Its output:
________________________
p0_xy = (-1, 0) projected to p0_x = (-1, 0)
p0_xy = (-1, 0.111111) projected to p0_x = (-0.777778, 0)
p0_xy = (-1, 0.222222) projected to p0_x = (-0.555556, 0)
p0_xy = (-1, 0.333333) projected to p0_x = (-0.333333, 0)
p0_xy = (-1, 0.444444) projected to p0_x = (-0.111111, 0)
p0_xy = (-1, 0.555556) projected to p0_x = (0.111111, 0)
p0_xy = (-1, 0.666667) projected to p0_x = (0.333333, 0)
p0_xy = (-1, 0.777778) projected to p0_x = (0.555556, 0)
p0_xy = (-1, 0.888889) projected to p0_x = (0.777778, 0)
p0_xy = (-1, 1) projected to p0_x = (1, 0)
________________________


A graphic:

https://i.ibb.co/3y5GzRq/ct-test-line.png


The radically simple sequence goes to 1, and is projected onto the unit
real line. Now, I can map more complex sequences using the same
technique. Btw, the circle in white is the unit circle, centered on {0,
0}, radius one.

What am I missing? Thanks.

[sergio]

x and y are orthagonal, so no problem with projection on to either axis

corrected i
i) The limit of {x[n]} is also the projection of the limit of {P[n]} onto the x axis


corrected ii
ii) The x-coordinate of the limit of {P[n]} is also the limit of {x[n]}



>
> --
> Rich

[Chris M. Thomasson]

On 1/21/2022 5:25 PM, Chris M. Thomasson wrote:
> On 1/20/2022 7:53 PM, RichD wrote:
>> Given an infinite sequence {P[n]}, in the x-y plane,
>> which converges to a limit point L.  Project those points
>> onto the x-axis, i.e. their x coordinates, creating the
>> sequence {x[n]}.  Finally, project L onto the x axis,
>> call this point M.
>>
>> Consider:
>> i)  M is the limit of {x[n]}
>> ii)  The limit of {x[n]} is M.
>>
>> Do both these assertions qualify as theorems?
>
> For some damn reason, this is making me thing of the following simple
> setup. Very quick, and I must be missing something important, anyway:
>

[...]

> Its output:
> ________________________
> p0_xy = (-1, 0) projected to p0_x = (-1, 0)
> p0_xy = (-1, 0.111111) projected to p0_x = (-0.777778, 0)
> p0_xy = (-1, 0.222222) projected to p0_x = (-0.555556, 0)
> p0_xy = (-1, 0.333333) projected to p0_x = (-0.333333, 0)
> p0_xy = (-1, 0.444444) projected to p0_x = (-0.111111, 0)
> p0_xy = (-1, 0.555556) projected to p0_x = (0.111111, 0)
> p0_xy = (-1, 0.666667) projected to p0_x = (0.333333, 0)
> p0_xy = (-1, 0.777778) projected to p0_x = (0.555556, 0)
> p0_xy = (-1, 0.888889) projected to p0_x = (0.777778, 0)
> p0_xy = (-1, 1) projected to p0_x = (1, 0)
> ________________________
>
>
> A graphic:
>
> https://i.ibb.co/3y5GzRq/ct-test-line.png
>
>
> The radically simple sequence goes to 1, and is projected onto the unit
> real line. Now, I can map more complex sequences using the same
> technique. Btw, the circle in white is the unit circle, centered on {0,
> 0}, radius one.
>
> What am I missing? Thanks.

https://i.ibb.co/TH6vT9N/ct-test-line-p1.png

Going from:
________________________

void manifest(ct::plot::cairo::plot_2d& plot)
{
std::cout << "ct_project_test::manifest\n\n";

{

glm::vec2 l0_p0 = { -1, -1 };

glm::vec2 l0_p1 = { -1, 1 };

glm::vec2 l1_p0 = { -1, -1 };
glm::vec2 l1_p1 = { 1, -1 };

unsigned long n = 10;

limit_line(plot, l0_p0, l0_p1, l1_p0, l1_p1, n);
}
}
________________________

output:
________________________
p0_xy = (-1, -1) projected to p0_x = (-1, -1)
p0_xy = (-1, -0.777778) projected to p0_x = (-0.777778, -1)
p0_xy = (-1, -0.555556) projected to p0_x = (-0.555556, -1)
p0_xy = (-1, -0.333333) projected to p0_x = (-0.333333, -1)
p0_xy = (-1, -0.111111) projected to p0_x = (-0.111111, -1)
p0_xy = (-1, 0.111111) projected to p0_x = (0.111111, -1)
p0_xy = (-1, 0.333333) projected to p0_x = (0.333333, -1)
p0_xy = (-1, 0.555556) projected to p0_x = (0.555556, -1)
p0_xy = (-1, 0.777778) projected to p0_x = (0.777778, -1)
p0_xy = (-1, 1) projected to p0_x = (1, -1)
________________________

Graphic:

https://i.ibb.co/TH6vT9N/ct-test-line-p1.png

[Chris M. Thomasson]

On 1/21/2022 6:07 PM, sergio wrote:
> On 1/21/2022 5:33 PM, RichD wrote:
>> On January 20, RichD wrote:
>>> Given an infinite sequence {P[n]}, in the x-y plane,
>>> which converges to a limit point L. Project those points
>>> onto the x-axis, i.e. their x coordinates, creating the
>>> sequence {x[n]}. Finally, project L onto the x axis,
>>> call this point M.
>>>
>>> Consider:
>>> i) M is the limit of {x[n]}
>>> ii) The limit of {x[n]} is M.
>>
>> hmmm.. trying to present a tricky problem, I flubbed the phrasing.
>> It's defective.
>>
>> Re-try:
>> Given an infinite sequence {P[n]}, in the x-y plane.  Project those
>> points
>> onto the x-axis, i.e. their x coordinates, creating the sequence {x[n]}
>
>>
>> Now:
>> i)  The limit of {x[n]} is also the projection of the limit of {P[n]}
>> ii)  The x-projection of the limit of {P[n]} is also the limit of {x[n]}
>>
>> True/false/maybe? Counterexamples?
>>
>
> x and y are orthagonal, so no problem with projection on to either axis

What type of projection? No problem with a conformal map:

https://youtu.be/Y6wOS7c7_Xg

Even preserves circles and ellipsoids...

https://youtu.be/bpBvK-VhSjA

;^)

[sergio]

those are excellent projectors, for both x and y, and the z, the second is has blinking ellopsoids

[RichD]

On January 21, Mike Terry wrote:
>> Re-try:
>> Given an infinite sequence {P[n]}, in the x-y plane. Project those points
>> onto the x-axis, i.e. their x coordinates, creating the sequence {x[n]}
>> Now:
>> i) The limit of {x[n]} is also the projection of the limit of {P[n]}
>> ii) The x-projection of the limit of {P[n]} is also the limit of {x[n]}
>> True/false/maybe? Counterexamples?
>
> They both still say the same thing, because saying A=B is always
> equivalent to saying B=A.

In mathematics, yes, but not in plain English.
One of the traps in this problem -

> You've dropped the requirement that P[n] converges, which changes things a bit.

That's deliberate.

> Let's call the x- and y-projection functions f_x and f_y respectively...
> IF P[n] CONVERGES (in the plane) then BOTH f_x(P[n]) and f_y(P[n]) will converge, and
> lim [ f_x( P[n] ) ] = f_x( lim [ P[n] ] ) [A]

That's (ii) above.
So you agree that (ii) is a theorem, assuming P[n] converges?

> Conversely, if BOTH x[n] and y[n] converge, then P[n] = (x[n], y[n]) will converge, and
> lim [ P[n] ] = lim [ (x[n],y[n]) =
> = ( lim [x[n]], lim [y[n] ) [C]

In this case, we ignore y[n].
You claim that (i) is a theorem, i.e. always true, when x[n] converges?

> Those results can be proved from the definition of convergence in the plane R^2,
> and convergence in R. (If you're familiar with metric spaces, then there is just one
> definition of "convergence" for metric spaces, but in your scenario we have two metric spaces
> involved, R and R^2, each with their own metric defined.)
>
> If a pre-condition above does not hold, the conclusion will not necessarily hold - e.g. if P[n] does
> NOT converge, it would still be possible for f_x(P[n]) to converge, but if it does then we could be
> sure f_y(P[n]) does not converge, and vice-versa swapping x,y. E.g. when P[n] = ( 0, n ) obviously
> f_x( P[n] ) = 0, which converges to 0, but f_x( P[n] ) = n which does not converge...
> (alternatively, maybe /neither/ projected sequence converges, as when P[n] = ( n, n ).)

So (i) isn't a theorem?
Counter-example?

> The projection functions f_x and f_y are /continuous/ functions between R^2 and R, and the above
> results are consequences of this.
> Hmm, perhaps there is confusion on the use of the term "the limit of..."? We don't use that term,
> or the expression "lim [ P[n] ]" unless we know that the sequence involved converges...
> (That's why I was careful in my wording for the results above, e.g. not writing [A]
> until after I've stated that both P[n] and f_x(P[n]) converge. An expression like
> f_x( lim [ P[n] ] ) is meaningless when P[n]
> does not converge, and so shouldn't even be uttered! :) )

(i) and (ii) are theorems, assuming appropriate conditions.

--
Rich

[Keegan Buker]

RichD wrote:

>> They both still say the same thing, because saying A=B is always
>> equivalent to saying B=A.
>
> In mathematics, yes, but not in plain English.
> One of the traps in this problem -

bullshit.

both in math stays for the value of RHS is assigned LHS.

[Mike Terry]

On 22/01/2022 20:45, RichD wrote:
> On January 21, Mike Terry wrote:
>>> Re-try:
>>> Given an infinite sequence {P[n]}, in the x-y plane. Project those points
>>> onto the x-axis, i.e. their x coordinates, creating the sequence {x[n]}
>>> Now:
>>> i) The limit of {x[n]} is also the projection of the limit of {P[n]}
>>> ii) The x-projection of the limit of {P[n]} is also the limit of {x[n]}
>>> True/false/maybe? Counterexamples?
>>
>> They both still say the same thing, because saying A=B is always
>> equivalent to saying B=A.
>
> In mathematics, yes, but not in plain English.
> One of the traps in this problem -

So perhaps you should give the mathematical equivalents of (i) and (ii), and we can see how they are
different. Note: this is sci.math, so people expect mathematical statements to have their
mathematical meaning, not their "plain English according to RichD" meaning!

Personally, I think you're just confused...

>
>> You've dropped the requirement that P[n] converges, which changes things a bit.
>
> That's deliberate.
>
>> Let's call the x- and y-projection functions f_x and f_y respectively...
>> IF P[n] CONVERGES (in the plane) then BOTH f_x(P[n]) and f_y(P[n]) will converge, and
>> lim [ f_x( P[n] ) ] = f_x( lim [ P[n] ] ) [A]
>
> That's (ii) above.
> So you agree that (ii) is a theorem, assuming P[n] converges?

Yes. (x[n] will always converge, so both "lim [ f_x( P[n] ) ]" and
"f_x( lim [ P[n] ] )" are well defined, and they are equal.

i.e. lim [ f_x( P[n] ) ] = f_x( lim [ P[n] ] )

Additionally, repeating what I said above, assuming P[n] converges we also have:

a) "the limit of {x[n]}" = lim [ f_x( P[n] ) ]
b) "the projection of the limit of {P[n]}" = f_x( lim [ P[n] ] )
c) So the limit of {x[n]} is also the projection of the limit of {P[n]}.

So your (i) above is true too. (Unsurprisingly.)

>
>> Conversely, if BOTH x[n] and y[n] converge, then P[n] = (x[n], y[n]) will converge, and
>> lim [ P[n] ] = lim [ (x[n],y[n]) =
>> = ( lim [x[n]], lim [y[n] ) [C]
>
> In this case, we ignore y[n].

No, I said y[n] converges. That's not ignoring y[n].

> You claim that (i) is a theorem, i.e. always true, when x[n] converges?

As I said above, (i) is a theorem if P[n] converges. If P[n] has no limit, then "the projection of
the limit of {P[n]}" is a nonsense phrase so neither (i) nor (ii) make sense. Also, it should be
clear that x[n] converging does not imply that P[n] converges.

Well that's using the mathematical understanding of what (i) says rather than your "plain English"
interpretation, which seems to disagree with my plain English interpretation.

You need to clarify what you think (i) means "mathematically"...

[RichD]

On January 22, Mike Terry wrote:
>>>> Re-try:
>>>> Given an infinite sequence {P[n]}, in the x-y plane. Project those points
>>>> onto the x-axis, i.e. their x coordinates, creating the sequence {x[n]}
> >>> Now:
>>>> i) The limit of {x[n]} is also the projection of the limit of {P[n]}
>>>> ii) The x-projection of the limit of {P[n]} is also the limit of {x[n]}
>>>> True/false/maybe? Counterexamples?
>
>>> They both still say the same thing, because saying A=B is always
>>> equivalent to saying B=A.

A kangaroo is a mammal.
Therefore, a mammal is a kangaroo.
?
English is not math.

>> In mathematics, yes, but not in plain English.
>> One of the traps in this problem -
>> So perhaps you should give the mathematical equivalents of (i) and (ii), and we can see
>> how they are different.

It isn't too difficult: {P[n]} doesn't necessarily converge - its
convergence isn't stipulated - in the x-y plane, while {x[n]},
the projection sequence, might still converge.
So (i) might be false.
ex. y = 1/(3-x)

But (ii) assumes {P[n]} does converge, and is always true.

>>> You've dropped the requirement that P[n] converges, which changes things a bit.
>
>> That's deliberate.
>
>>> Let's call the x- and y-projection functions f_x and f_y respectively...
>>> IF P[n] CONVERGES (in the plane) then BOTH f_x(P[n]) and f_y(P[n]) will converge, and
>>> lim [ f_x( P[n] ) ] = f_x( lim [ P[n] ] ) [A]
>
>> That's (ii) above.
>> So you agree that (ii) is a theorem, assuming P[n] converges?
>
> Yes. (x[n] will always converge, so both "lim [ f_x( P[n] ) ]" and
> "f_x( lim [ P[n] ] )" are well defined, and they are equal.
> i.e. lim [ f_x( P[n] ) ] = f_x( lim [ P[n] ] )
>
> Additionally, repeating what I said above, assuming P[n] converges we also have:
> a) "the limit of {x[n]}" = lim [ f_x( P[n] ) ]
> b) "the projection of the limit of {P[n]}" = f_x( lim [ P[n] ] )
> c) So the limit of {x[n]} is also the projection of the limit of {P[n]}.
> So your (i) above is true too. (Unsurprisingly.)
> >
> >> Conversely, if BOTH x[n] and y[n] converge, then P[n] = (x[n], y[n]) will converge, and
> >> lim [ P[n] ] = lim [ (x[n],y[n]) =
> >> = ( lim [x[n]], lim [y[n] ) [C]
>
> > In this case, we ignore y[n].
> No, I said y[n] converges. That's not ignoring y[n].

The problem says nothing about y[n].

>> > You claim that (i) is a theorem, i.e. always true, when x[n] converges?
>
> As I said above, (i) is a theorem if P[n] converges. If P[n] has no limit, then "the projection of
> the limit of {P[n]}" is a nonsense phrase so neither (i) nor (ii) make sense. Also, it should be
> clear that x[n] converging does not imply that P[n] converges.

Right. That's the main thrust of the problem.

>>> Those results can be proved from the definition of convergence in the plane R^2,
>>> and convergence in R. (If you're familiar with metric spaces, then there is just one
>>> definition of "convergence" for metric spaces, but in your scenario we have two metric spaces
>>> involved, R and R^2, each with their own metric defined.)

To be precise, we should specify the Euclidean metric R^2 for the plane.
Usually that's assumed.

> > So (i) isn't a theorem?
> > Counter-example?
> >
>>> The projection functions f_x and f_y are /continuous/ functions between R^2 and R, and the above
>>> results are consequences of this.

Is the projection function always continuous?
I used to know this stuff, but the neurons rust -


--
Rich

[Mike Terry]

On 23/01/2022 23:27, RichD wrote:
> On January 22, Mike Terry wrote:
>>>>> Re-try:
>>>>> Given an infinite sequence {P[n]}, in the x-y plane. Project those points
>>>>> onto the x-axis, i.e. their x coordinates, creating the sequence {x[n]}
>>>>> Now:
>>>>> i) The limit of {x[n]} is also the projection of the limit of {P[n]}
>>>>> ii) The x-projection of the limit of {P[n]} is also the limit of {x[n]}
>>>>> True/false/maybe? Counterexamples?
>>
>>>> They both still say the same thing, because saying A=B is always
>>>> equivalent to saying B=A.
>
> A kangaroo is a mammal.
> Therefore, a mammal is a kangaroo.
> ?
> English is not math.
>

But your (i) and (ii) are "is" relationships (suggesting equals) and certainly not "is a"
relationships. Anyhow, I think I get what you're saying from your comments below.

Here is how I would have said them:

i) If x[n] converges, does P[n] necessarily converge, and if so is
lim x[n] = proj_x (lim P[n]) ?
ii) If P[n] converges, does x[n] necessarily converge, and if so is
lim x[n] = proj_x (lim P[n]) ?

Seems clearer to me, but maybe that's just me...

All linear maps (for finite dimensional vector spaces over R at least) are continuous. So that
covers projections, and lots more.

> I used to know this stuff, but the neurons rust -

Yeah, you've reminded me that I've forgotten a mountain of stuff around topological vector spaces
and functional analysis. (Infinite dimensional spaces and so on...) Well that was 40-ish years
ago, so no surprise I guess.) :)

Mike.

[Barry Schwarz]

On Sun, 23 Jan 2022 15:27:13 -0800 (PST), RichD
<r_dela...@yahoo.com> wrote:

>On January 22, Mike Terry wrote:
>>>>> Re-try:
>>>>> Given an infinite sequence {P[n]}, in the x-y plane. Project those points
>>>>> onto the x-axis, i.e. their x coordinates, creating the sequence {x[n]}
>> >>> Now:
>>>>> i) The limit of {x[n]} is also the projection of the limit of {P[n]}
>>>>> ii) The x-projection of the limit of {P[n]} is also the limit of {x[n]}
>>>>> True/false/maybe? Counterexamples?
>>
>>>> They both still say the same thing, because saying A=B is always
>>>> equivalent to saying B=A.
>
>A kangaroo is a mammal.
>Therefore, a mammal is a kangaroo.
>?
>English is not math.

So {P[n]} is an infinite sequence.
Therefore, any infinite sequence must equal {P[n]}

Bad logic is bad logic, regardless of language.