0
Let (an) be a convergent sequence in metric space X. Let "a" be the limit of an. Then for sufficiently large N, if x,y>N, d(ay,a)<e/2 and d(ax,a)<e/2 for any e greater than zero.
We add these inequalities and get d(ay,a) + d(ax,a)<e. But by the triangle inequality d(ay,ax)<d(ay,a) + d(ax,a), so
d(ay,ax)<e for all x,y>N.
(some of those < should be less than or equal to, but it doesn't really affect the proof.)
What else can I help you with?
How do you prove that the sum of a convergent sequence divided by n will converge?
You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.
Is compact metric space is complete?
A compact metric space is not necessarily complete. Compactness only guarantees that every sequence in the space has a convergent subsequence, while completeness requires that every Cauchy sequence converges to a point in the space.
Prove that every convergent sequence is a Cauchy sequence?
The limits on an as n goes to infinity is aThen for some epsilon greater than 0, chose N such that for n>Nwe have |an-a| < epsilon.Now if m and n are > N we have |an-am|=|(am -a)-(an -a)|< or= |am -an | which is < or equal to 2 epsilor so the sequence is Cauchy.
Prove that countable space is countable?
prove that every metric space is hausdorff and first countable
Why must every proposition in an argument be tested using a logical sequence?
A logical sequence in an argument is a way to prove a step has a logical consequence. Every proposition in an argument must be tested in this fashion to prove that every action has a reaction.
Given an arbitrary sequence of digits how can you prove it appears somewhere in pi?
It has not yet been proven whether any arbitrary sequence of digits appears somewhere in the decimal expansion of pi.
How do you prove that if a real sequence is bounded and monotone it converges?
We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.Now since {an} is increasing for all n greater than N we have |S-an|
How do you prove the shortest line between two points is the segment that joins them on a plane?
The shortest line between two points is NOT always the segment that joins (or jion) them on a plane: the answer depends on the concept of distance or the metric used for the space. If using a taxicab or Manhattan metric it is the sum of the North-South distance and the East-West distance. There are many other possible metrics.The proof for a general metric is the Cauchy-Schwartz inequality but this site is totally incapable of dealing with the mathematical symbols required to prove it.
Prove that Hilbert Space is a Metric Space?
The question doesn't make sense, or alternatively it is true by definition. A Hilbert Space is a complete inner product space - complete in the metric induced by the norm defined by the inner product over the space. In other words an inner product space is a vector space with an inner product defined on it. An inner product then defines a norm on the space, and every norm on a space induces a metric. A Hilbert Space is thus also a complete metric space, simply where the metric is induced by the inner product.
How can you prove to anyone in doubt that US automobiles built since 1975 are metric in engineering and design and manufacture and service?
At the time our president (Carter?) agreed to convert to the metric system if the other developed countries (Europe) would conduct business using the American Dollar as the standard for trade. You see how that worked out.
How do you prove 1 kg equal to 1000 grams?
To prove that 1 kg is equal to 1000 grams, you can use the definition of the kilogram as the base unit of mass in the metric system. By definition, 1 kilogram is equal to 1000 grams. This relationship is established and accepted in the International System of Units (SI).
Do the theorems for junior certificate maths have to be exactly right or do they just have to make sense?
There is more than one way to prove a given mathematical proposition. If the sequence of reasoning is valid, then the proof is correct. That is all that is required.
