rational sphere

Originally posted by Palynka
But does then a photon have volume? I'm confused 😕

Sort of. Microwaves are apparently too big to fit through the wire mesh in the door of the microwave; they're ~1cm around apparently.

Originally posted by Palynka
To disprove it, all you need is to construct a sequence of numbers within the set that converges to a number outside the set. So if you pick any sequence that converges to say (sqrt(1/3),sqrt(1/3),sqrt(1/3)), then you disprove local compactness. Since the set is dense (using your proof) we know that we can get arbitrarily close to any point on the real sphere so such a sequence exists.

No, that proves denseness.

A subset X of a space Y is called dense in Y if:

For every point y in Y there is a sequence x1, x2, x3, ... in X so that
For all epsilon there is a natural number N so that for all m > N
|xm - y| < epsilon

Compactness of a space has nothing to do with points outside that space.

Originally posted by TheMaster37
No, that proves denseness.

A subset X of a space Y is called dense in Y if:

For every point y in Y there is a sequence x1, x2, x3, ... in X so that
For all epsilon there is a natural number N so that for all m > N
|xm - y| < epsilon

Compactness of a space has nothing to do with points outside that space.

Like you say denseness means that for each point of the space there is at least one sequence like the one you describe.

Compactness can be disproven by showing that there is at least one sequence that converges to a point outside the space.

These are different things. Now you have me doubting if that disproves compactness, but what I said was definitely not about denseness.

Originally posted by Palynka
Like you say denseness means that for each point of the space there is at least one sequence like the one you describe.

Compactness can be disproven by showing that there is at least one sequence that converges to a point outside the space.

These are different things. Now you have me doubting if that disproves compactness, but what I said was definitely not about denseness.

I've digged up my old topology book and found the definition we used:

A topological space X is called compact if for every arbritary collection Ua (a in A) of open subsets, so that the union of all Ua (a in A) = X,
there exists a finite subset J of A so that the union of all Ua (a in J) = X

From this follows tat any closed and bounded space is compact, whereas a open space is not compact.

With that it appears you were right, if there is a sequence converging to a point outisde your space (meaning the space is open) cannot be compact.