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### Closure and Compactness in a Metric Space

```Date: 10/08/2002 at 20:24:03
From: Pham
Subject: Compactness/ closed/open

Regard Q, the set of all rational numbers, as a metric space, with
d(p,q)=|p-q|. Let E be the set of all P in Q such that 2<P^2<3. Show
that E is closed and bounded in Q, but that E is not compact.
Is E open in Q?

Thank you very much.
```

```
Date: 11/20/2002 at 23:21:58
From: Doctor Nitrogen
Subject: Re: Compactness/ closed/open

Hi, Pham:

Try to see if this reasoning helps.

Find convergent sequences of rational numbers

P = 1 + a_n, n = 1, 2, 3, ........,

with a_n = m/n, m, n integers, and m/n always an element of Q such
that

sqrt(2) < 1 + a_n < sqrt(3),

and, for P = (1 + a_n) and P^2 = (1 + a_n)^2 also both converging:

2 < (1 + a_n)^2 < 3 for each n.

Then the set

E:= {All P in Q | 2 < P^2 < 3}

can be shown to exist.

Then E can be shown to contain all its limit points, meaning E must
be closed in Q. You will, of course, have to find suitable candidates
for a_n and m/n.

But also, this set E of elements P, with P = 1 + a_n, would have both
P and also P^2 bounded below and above, so E would be bounded in Q.
So E is both closed and bounded in Q.

But what about E when viewed as a METRIC SPACE in Q? Is it a compact
metric space in Q with d(p, q) = |p - q|?

Let p and q both be elements satisfying the definition for E, such
that

p = 1 + a_n,

q = 1 + b_n, n = 1, 2, 3, ....,

and such that p^2 and q^2 are both convergent sequences with limits
inside(2, 3), but for which

d(p, q) = |p - q| = sqrt(2), that is,

lim n --> oo |(1 + a_n) - (1 + b_n)|

= lim n --> oo |a_n - b_n| = sqrt(2).

If you can find two such sequences p = 1 + a_n and q = 1 + b_n, maybe
so that

(1 + a_1) - (1 + b_1) = 1.0,

(1 + a_2) - (1 + b_2) = 1.4,

(1 + a_3) - (1 + b_3) = 1.41,

(1 + a_4) - (1 + b_4) = 1.414,

.............................., etc.,

just to illustrate,

this would mean there exists a Cauchy sequence that does not converge
either in E nor in Q, but in R (reals), meaning E cannot be a complete
metric space inside Q. But this implies E cannot be "sequentially
compact" inside Q (you might want to do some research on this term),
meaning E cannot be compact in Q.

Is every point inside E interior to E?

If every point of E is interior to E, E is open in Q. If the elements
of E are not all interior to E, then E is closed in Q.

One other alternative to prove E is not compact in Q might be to find
an open covering of E which has no finite subcovering for E.

- Doctor Nitrogen, The Math Forum
http://mathforum.org/dr.math/
```
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College Analysis
High School Analysis
High School Sets

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