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Compact Observation
if atom would have been a compact mass, what would be the observations of rutherford experiment?
If atom would have been a compact mass, what would be the observations of rutherford experiment
The tiny positively charged particles, traveled from a positively charged metal disc to a very thin piece of Gold foil. A few of the particles were slightly deflected, as if they hit edge of a very dense object. Even fewer of the particles were deflected 180°, so these particles returned straight back to the source, as if they had hit the center of a very dense, positively charged object.
Most of the particles went straight through the gold foil, as if the gold foil full of holes. Later they discovered the gold foil and all matter is full of mostly nothing. The nucleus occupies approximately 1/10,000th of the volume of an atom. Nothing occupies 1/9,999th of the volume of an atom. We call the nothing, a vacuum!
If the atoms of gold would have been a compact mass, the evidence would be reversed. Maaaaby a feeeeew of the particles might find a way to move through the gold foil. But most likely, 99.9999% of the particles would be reflected back toward the source at various angles.
As a result, no one would ever have wondered what tiny negatively charged particles might be moving in the empty space,vacuum. No one would be looking for electrons. NO ONE LOOKING MEANS NO ONE DISCOVERING!!
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Pointwise limit help?
Assume fn → f pointwise on a compact set K and assume that for each x ∈ K the sequence fn(x) is increasing. Follow these steps to show that if fn and f are continuous on K, then the convergence is uniform.
a. set gn = f - fn and translate the preceding hypothesis into statements about the sequence (gn)
b. let e > 0 be arbitrary, and define Kn = { x ∈ K : gn(x) ≥ e}. Argue that K1 ⊇ K2 ⊇ K3 ⊇ .... is a nested sequence of compact sets, and use this observation to finish the argument.
Note to poster: I hope this solution is worth the wait!
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a) Let gn = f - fn.
Note that gn = f - fn → f - f = 0
and fn(x) >= fn(y) <==> f - fn(x) <= f - fn(y).
Moreover, since (fn(x)) is increasing with x fixed, we see that g(x) ≥ 0 for all x ∈ K.
The restatement in terms of (gn) is as follows:
Assume gn → 0 pointwise on a compact set K and assume that for each x ∈ K the sequence gn(x) is decreasing. If gn are continuous on K, then the convergence is uniform.
b) Let ɛ > 0 be given and define Kn = {x ∈ K : gn(x) ≥ ɛ}.
Since Kn is a closed subset of a compact set K, Kn itself is compact (for each n).
Next, Ki ⊇ Kj for any i < j, because for each x ∈ K the sequence gn(x) is decreasing.
So, we may apply the compact set generalization of the Nested Interval Property to conclude that the intersection of all K1, K2, ... is not empty; let's call this element z.
Thus, gn(z) ≥ ɛ for all n, which contradicts gn(z) → 0 as n→ infinity.
Therefore, there exists a positive integer N such that for all x in K
gn(x) < ɛ for all n >= N.
In other words, given ɛ > 0 there exists a positive integer N such that
for all n >= N and x ∈ K, |gn(x) - 0| = gn(x) < ɛ.
Hence, gn → 0 uniformly on K.
I hope this helps!
