Proving Compactness of Projected Sets Using Sequences and Subsequences

[kidmode01]

Say there is a sequence of points: {$$x_k,y_k$$} that has a convergent subsequence:

{$${x_k_i,y_k_i}}$$} that converges to: $$(x_0,y_0)$$.

Sorry for poor latex, it should read "x sub k sub i"

Can I extrapolate the sequence {$$x_k_i$$} and say it converges to $$x_0$$ seperately?

The reason I ask this is because I would like to show that the projection of a compact set S in the x,y plane to the x-axis is also compact. Basically picking a sequence $$x_k$$ in the projection , finding a corresponding sequence {$$x_k,y_k$$} in S where $$y_k$$ is arbitrary, that has a convergent subsequence whose limit is $$(x_0,y_0)$$, but then if I can bust that subsequence apart I can show the sequence in the projection has a convergent subsequence thus proving compactness (since sequentially compactness implies compactness for subsets of R^n)

Or do I need to project the subsequence in S down to the x-axis first? It seems like kind of "hand waving math" to just pull apart the subsequence and say each sequence of coordinates converges to a particular coordinate. Could someone point me in the right direction?

 

[CompuChip]

I think you can do that.
I even think you can make it rigorous by applying a projection on the first coordinate: P(x, y) = x: if you show that the projection is continuous then you can take the limit of the projection of the sequence and then swap the limit and projection, because for continuous functions f(x),
$$\lim_{x \to a} f(x) = f( \lim_{x \to a} x )$$

OK, I know it's not perfect, but it's a start

 

[kidmode01]

Well there is a theorem that states the image of a continuous function whose domain is a compact set is also compact but I didn't want to use any continuity for this proof. But you I know what you mean.

I think for my question I can say specifcally the x_k_i's converge to x0 and the y_k_i's converge to y0 just by the definition of a convergent sequence in R^n

 

[CompuChip]

Yeah, so the conclusion is that the "hand-waving math" is right, and that it is possible to prove if you like.