Showing Convergent Subsequence Exists

In summary, the problem states that in the space ##([0, 1], d_1)##, there exists a sequence ##(x_n)## where for any ##x \in [0, 1]##, there exists a subsequence ##(x_{n_k})## that converges to ##x##. This may seem contradictory to a previous theorem, but in reality, it just means that the main sequence has multiple limit points and thus, multiple subsequences converging to different limits. The solution to this problem is surprisingly simple.
  • #1
transmini
81
1

Homework Statement


Consider the space ##([0, 1], d_1)## where ##d_1(x, y) = |x-y|##. Show that there exists a sequence ##(x_n)## in ##X## such that for every ##x \epsilon [0, 1]## there exists a subsequence ##(x_{n_k})## such that ##\lim{k\to\infty}\space x_{n_k} = x##.

Homework Equations


N/A

The Attempt at a Solution


We had no idea where to even begin with this one. It doesn't seem to make sense in the first place when you take the theorem:

Let ##(x_n)## be sequence, and let ##x \epsilon X##. Prove that if ##\lim{n\to\infty}\space x_n = x##, then for every subsequence ##(x_{n_k})## we have ##\lim{k\to\infty} \space x_{n_k} = x##

With the problem saying that there is specific sequence such that for any limit point we can find a subsequence, this implies there a multiple subsequences with different limits, which contradicts what this previous theorem says.

Any suggestions on where to begin with this one?
 
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  • #2
transmini said:

Homework Statement


Consider the space ##([0, 1], d_1)## where ##d_1(x, y) = |x-y|##. Show that there exists a sequence ##(x_n)## in ##X## such that for every ##x \epsilon [0, 1]## there exists a subsequence ##(x_{n_k})## such that ##\lim{k\to\infty}\space x_{n_k} = x##.

Homework Equations


N/A

The Attempt at a Solution


We had no idea where to even begin with this one. It doesn't seem to make sense in the first place when you take the theorem:

Let ##(x_n)## be sequence, and let ##x \epsilon X##. Prove that if ##\lim{n\to\infty}\space x_n = x##, then for every subsequence ##(x_{n_k})## we have ##\lim{k\to\infty} \space x_{n_k} = x##

With the problem saying that there is specific sequence such that for any limit point we can find a subsequence, this implies there a multiple subsequences with different limits, which contradicts what this previous theorem says.

Any suggestions on where to begin with this one?

There are no contradictions in the problem statement. The main sequence ##\{x_n\}## does not have a limit; it has numerous limit points, meaning that it has numerous subsequences converging to their own, individual limits. The problem asserts that there is a sequence such that any ##x \in [0,1]## is a limit of some subsequence. Believe it or not, there is a very easy solution.
 

Related to Showing Convergent Subsequence Exists

1. What is meant by "convergent subsequence"?

A convergent subsequence is a part of a sequence that approaches a specific number or value as the number of terms in the subsequence increases. In other words, the terms in the subsequence get closer and closer to a particular value as more terms are added.

2. Why is it important to show that a convergent subsequence exists?

Showing that a convergent subsequence exists is important in proving the convergence of a sequence. It allows us to determine the limit of the sequence and understand how the terms in the sequence behave as the number of terms increases. This information is crucial in many mathematical and scientific applications.

3. How do you prove the existence of a convergent subsequence?

To prove the existence of a convergent subsequence, we need to use the Bolzano-Weierstrass theorem. This theorem states that every bounded sequence has a convergent subsequence. This means that if we can show that a sequence is bounded, then we can prove the existence of a convergent subsequence.

4. What are some common methods used to show the existence of a convergent subsequence?

One common method is to use the monotone convergence theorem, which states that if a sequence is monotone (either increasing or decreasing) and bounded, then it must converge. Another method is to use the Cauchy criterion, which states that a sequence converges if and only if it is a Cauchy sequence (i.e. the terms get arbitrarily close to each other as the number of terms increases).

5. Are there situations where a convergent subsequence may not exist?

Yes, if a sequence is unbounded or does not have a limit, then it may not have a convergent subsequence. Additionally, if a sequence is oscillating or does not follow a clear pattern, it may not have a convergent subsequence. In these cases, alternative methods may need to be used to analyze the behavior of the sequence.

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