Hello! I have 4 questions below that I would appreciate some help on. Please sho

Question

Hello! I have 4 questions below that I would appreciate some help on. Please show work.

1. Let (X,d) be a metric space where d is the discrete metric. Suppose that {Xn} is a convergent sequence X. Show that there exists a K in N such that for all n >= K we have Xn=Xk.

2. Let d(x,y) = min{1, lx-yl}.

   a.) Show that d defines a metric on R.

   b.) Show that a sequence converges in (R,d) if and only if it converges in the standard metric.

   c.) Find a bounded sequence in (R,d) that contains no convergent subsequence.

3. Suppose that {Vn} from n=1 to infinity, is a collection of open sets in a metric space (X,d) such that Vn is a subset of Vn+1. Let {Xn} from n=1 to infinity, be a sequence such that Xn in Vn+1 Vn   ( the means subraction) and suppose that {Xn} converges to p in X. Show that p in the boundary of V where V= U from n=1 to infinity of Vn.

The last question deals with compactness and comleteness:

4. Show that a compact set K is a complete metric space.

Explanation / Answer

Hope this helps:

b)) Let x, y ? X. Since d(x, y) ? 0, d(x, y) = min(d(x, y),1) ? 0, and

d(x, x) = min(d(x, x),1) = min(0,1) = 0.

If d(x, y) = min(d(x, y),1) = 0, then d(x, y) = 0, so x = y. This shows that the ?rst metric axiom holds.

Since d(x, y) = min(d(x, y),1) = min(d(y, x),1) = d(y, x), the second metric axiom holds.

To verify the third axiom, take x, y, z ? X. Begin by noting that if either d(x, y) ? 1 or d(y, z) ? 1,

then d(x, y) = 1 or d(y, z) = 1. Therefore,

d(x, y) + d(y, z) ? 1 ? d(x, z).

Assume that d(x, y) < 1 and d(y, z) < 1. Then

d(x, y) + d(y, z) = d(x, y) + d(y, z) ? d(x, z) ? d(x, z).

This veri?es the third axiom, so d is a metric.

4.We argue contrapositively. First we show that if a set K is not closed,

then it can not be compact, and then we show that if K is not bounded, it

can not be compact.

Assume that K is not closed. Then there is a boundary point a that does

not belong to K. For each n ? N, there is an xn ? K such that d(xn, a) <

1

n

.

The sequence {xn} converges to a /? K, and so does all its subsequences,

and hence no subsequence can converge to a point in K.

Assume now that K is not bounded. For every n ? N there is an element

xn ? K such that d(xn, b) > n. If {yk} is a subsequence of xn, clearly

limk?? d(yk, b) = ?. It is easy to see that {yk} can not converge to any

element y ? X: According to the Triangle Inequality

d(yk, b) ? d(yk, y) + d(y, b)

and since d(yk, b) ? ?, we must have d(yk, y) ? ?. Hence {xn} has no

convergent subsequences, and K can not be compact.

Hence, proved

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.