Sequences & Subsequences : IIT 2018 Problem 10

What are we learning?

Sequences, Subsequences are the key features in the field of real analysis. We will see how to imply these concepts in our problem

Understand the problem

Let \(s_n\) = 1+\(\frac{1}{1!}\)+\(\frac{1}{2!}\)+........+\(\frac{1}{n!}\) for n \(\in\) \(\mathbb{N}\) Then which of the following is TRUE for the sequence $\{s_{n}\}^\infty_{n=1}$:   (a) $\{s_{n}\}^\infty_{n=1}$ converges in $(\mathbb{Q})$ .   (b) $\{s_{n}\}^\infty_{n=1}$ is a Cauchy sequence but does not converges to $(\mathbb{Q})$.   (c) The subsequence  $\{s_{k^n}\}^\infty_{n=1}$ is convergent in $(\mathbb{R})$ when k is a even natural number.   (d) $\{s_{n}\}^\infty_{n=1}$ is not a Cauchy sequence. Difficulty Level Easy Suggested Book

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2 Tom M. Apostol

Start with hints

I am going to give you 3 clues in the beginning you try to work out using them. Then I will elaborate this clues in the following hints  (I) Every convergent sequence is a Cauchy sequence  (II)Every subsequence of a convergent sequence is convergent  (III)Consider then term 1+\(\frac{1}{1!}\)+\(\frac{1}{2!}\)+........+\(\frac{1}{n!}\) Does this remind you any well known series?

I wil start with (III) consider \(e^x\)=1+\(\frac{x}{1!}\)+\(\frac{x^2}{2!}\)+........+\(\frac{x^n}{n!}\) Isn't the seris that we have to , is the value at x=1. Hence the given series\(\rightarrow\) e \(\in\) \(\mathbb{R}\) \ \(\mathbb{Q}\)

So option (a) is incorrect.

Every subsequence of a convergent sequence is convergent so $\{s_{k^n}\}^\infty_{n=1}$ is convergent not only for even k, but for any \(k \in \Bbb N\). So option (c) is incorrect.

Every convergent sequence is a Cauchy sequence so option (d) is incorrect and \(e \in\) \(\mathbb{R}\) so the given subsequence is convergent in \(\mathbb{R}\). So only option (b) is correct.

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Problem:

Let $f: X\to Y$ be a continuous map between metric spaces. Then $f(X)$ is a complete subset of $Y$ if

A. X is compact

B. Y is compact

C. X is complete

D. Y is complete

Discussion:

Let $X$ be compact. Then $f(X)$ is compact. (continuous image of compact space is compact)

Now, compact subset of any Hausdorff space is closed. So in particular, compact subset of any metric space is closed.

Let $y_n$ be a Cauchy sequence in $f(X)$. Then since $f(X)$ is compact, $y_n$ has a convergent subsequence (converging to a point in that compact set i.e, in $f(X)$.

Suppose $y_{n_k} \to y \in f(X)$.

Then by triangle inequality, we have $d(y_n,y) \le d(y_n,y_{n_k}) + d(y_{n_k},y) \to 0+0=0$ as $n\to \infty$

Here we have used that $y_n$ is cauchy to conclude $d(y_n,y_{n_k}) \to 0$.

So this implies $y_n \to y$. Since $y\in f(X)$ we conclude that $f(X)$ is complete.

This proves A.

Let $Y=[0,2]$ and $X=(0,1)$. Take the inclusion map $i(x)=x$ for all $x\in X$. This example shows that even if we take $Y$ to be compact, or complete, $f(X)$ need not be complete. So this disproves B and D.

Now take $X=\mathbb{R}$ and take $Y=(0,1)$. We know there is a homeomorphism between these two sets where the metric is usual topology. So, in this case, the image of a complete set is not complete. This disproves option C.

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