Re: New Question in Metric Spaces

On 12 Oct, 06:54, miki <miki.li...@xxxxxxxxx> wrote:

[snip for bandwidth]


OK, now we're getting places. So it is true that there could be an
infinite set of such x_k's but it doesn't matter you say.
I only have to show that there is a finite set such that etc.. This is
well understood.

Good. Let's call that finite set F.


However, I thought that I have to show that

|g_n(x_k) - g_m(x_k)| < epsilon/3 for ALL x_k if m > n > n_max,
otherwise where is the uniformity . . .?

Now, you found some finite set x_k where 1<= k <=h and you took n_max
= max {n_k}. that's would definitely work for a sepcific finite set
you chose.
But if I got it right, I should find such an n value such that if m >
n > n_max

|g_n(x_k) - g_m(x_k)| < epsilon/3 for ALL x_k.

No, we only need to show it for x_k in F. The uniformity comes from
the fact that our sequence (g_n) is uniformly equicontinuous - that
is, for every epsilon > 0 there is a delta > 0 such that, for EVERY x,
y in X with rho(x,y) < delta and every g_n we have |g_n(x) - g_n(y)| <
epsilon/3. Recall that we have chosen such a delta, and that we have
chosen F in such a way that, for every x in X, rho(x,x_k) < delta for
some x_k in F. If n, m >= n_max then the fact that |g_n(x_k) - g_m
(x_k)| < epsilon/3 for all x_k in F is enough to prove that |g_n(x) -
g_m(x)| < epsilon for ALL x. For suppose rho(x,x_k) < delta - it
doesn't matter which x_k we choose, it only matters that we can choose
SOME x_k which is in the set F. Then

|g_n(x) - g_m(x)|
= |g_n(x) - g_n(x_k) + g_n(x_k) - g_m(x_k) + g_m(x_k) - g_m(x)|
<= |g_n(x) - g_n(x_k)| + |g_n(x_k) - g_m(x_k)| + |g_m(x_k) - g_m(x)|

(the last line follows from the triangle inequality). The term in the
middle is < epsilon/3, since x_k is in F. For the other two terms,
note that rho(x,x_k) < delta, and delta was chosen in such a way to
guarantee that both of these terms are < epsilon/3. So the whole thing
is < epsilon.
.

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