TIFR 2013 problem 25 | Complete metric on (0,1)

Try this problem from TIFR 2013 Problem 25 based on Complete Metric on (0,1).

Question: TIFR 2013 problem 25

True/False?

There exists a complete metric on the open interval \((0,1)\) inducing the usual topology.

Hint:

Topologically, (0,1) can be made "equal" to \(\mathbb{R}\), which is a complete space with usual metric.

Discussion:

Suppose \(f:(0,1)\to \mathbb{R} \) be a homeomorphism. (Which we know exists). Define a new distance function \(d\) on \((0,1)\) as follows:

for any \(x,y \in (0,1) \), \(d(x,y)=|f(x)-f(y)|\).

The fact that d is indeed a metric follows because we are essentially using Euclidean distance.

Hope: \( ((0,1),d) \) satisfies the condition of the statement.

Since \(f\) is a homeomorphism, the inverse function \(f^{-1}\) is continuous, which implies \(f\) takes open sets to open sets. And together with continuity, this implies that a set \(S\subset (0,1) \) is open in \((0,1)\) if and only if \(f(S)\) is open in \(\mathbb{R}\) which happens if and only if \(S\) is open in \(((0,1),d)\).

Therefore, \(S\subset (0,1) \) is open in \(0,1)\) (with respect to subspace topology) if and only if \(S\) is open in \(((0,1),d)\). This gives:

conclusion 1: \(((0,1),d)\) induces the usual topology.

Suppose \((x_n)\) is a Cauchy sequence in \(((0,1),d)\). That means, \(d(x_n,x_m) \to 0 \) as \(n,m \to \infty \). Which is same as \(|f(x_n)-f(x_m)| \to 0 \) as \(n,m \to \infty \). Now \((f(x_n))\) is a Cauchy sequence in \(\mathbb{R}\), therefore it has a limit \(y\) in \(\mathbb{R}\.

\(f(x_n) \to y \). By the continuity of \(f^{-1}\), \(x_n \to f^{-1}(y) \in (0,1) \). This gives:

conclusion 2:  \( ((0,1),d) \) is complete.

Remark: How do we know \((0,1)\) is homeomorphic with \(\mathbb{R}\)? Well there can be many homeomorphisms. Take any function which is "minus infinity" at 0 and "infinity" at 1. For example \(tan\) with some appropriate adjustments work. (Hint: shift \((0,1)\) to \( (-\pi /2,\pi /2) \) ).

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TIFR 2013 Problem 24 Solution -Non-existence of continuous function


TIFR 2013 Problem 24 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects. In general, TIFR entrance exam hits the floor during the month of December.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:True/False

There exists a continuous surjective function from \(S^1 \) onto \(\mathbb{R}\).

Hint:

Search for topological invariants.

Discussion:

We know that continuous image of a compact set is compact. \(S^1\) is a subset of \(\mathbb{R}^2\), and in \(\mathbb{R}^2\) a set is compact if and only if it is closed and bounded.

By definition, every element of \(S^1\) has unit modulus, so it is bounded.

Let's say \(z_n\to z\) as \(n\to \infty \). Where {\(z_n\)} is a sequence in \(S^1\). Since modulus is a continuous function, \(|z_n| \to |z| \), the sequence {\(|z_n|\)} is simply the constant sequence \(1,1,1,... \) hence \(|z|=1\).

What does above discussion mean? Well it means that if \(z\) is a limit point (or even a point of closure) of \(S^1\) then \(z\in S^1\).  Therefore, \(S^1\) is closed.

The immediate consequence is that the given statement is False. Because, \(\mathbb{R}\) is not compact. \(S^1\) is compact, and continuous image of a compact set has to be compact.

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Non-homeomorphic (TIFR 2013 Math Solution problem 22)

TIFR 2013 Math Solution Discussion

Question:

The sets \([0,1)\) and \((0,1)\) are homeomorphic.

Hint:

Check some topological invariant.

TIFR 2013 Math Solution

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Discussion:

In \([0,1)\), \(0\) seems to be a special point, as compared to \((0,1)\) where every point has equal importance (or non-importance).

If \(f:X\to Y \) is a homeomorphism then for any point \(a\in X\),  \(X- \{ a \} \) and \(Y- \{ f(a) \} \) are homeomorphic with the homeomorphism function being restriction of \(f\) to \(X- \{a \} \).

If \(f: [0,1) \to (0,1) \) is a homeomorphism, then choosing to remove \(0\) from \([0,1)\) we get that \( (0,1) \) is homeomorphic to \( (0,1)- \{f(0) \} \).

Whatever be \(f(0) \), we see that \( (0,1)- \{f(0) \} \) is a disconnected set. Whereas, \((0,1)\) is connected. A continuous image of a connected set is always connected. Hence we are forced to conclude that there was no such \(f\) to begin with.

So the statement is False.

TIFR 2013 problem 21 | No fixed point Homeomorphism

Try this problem from TIFR 2013 problem 21 based on no fixed homeomorphism.

Question: TIFR 2013 problem 21

True/False?

Every homeomorphism of the 2-sphere to itself has a fixed point.

Hint:

\(z= -z\) implies \(z=0\)

Discussion:

2-sphere means \( S^2=\left \{(x,y,z)\in\mathbb{R}^3 | x^2+y^2+z^2=1 \right \} \).

i.e, \( S^2=\left \{v\in\mathbb{R}^3 | ||v||=1 \right \} \).

\(||.||\) denotes the usual 2-norm (Euclidean norm).

Let us try \(f:S^2\to S^2\) defined by \(f(v)=-v\) for all \(v\in\mathbb{R}^3\).

The only vector in \(\mathbb{R}^3\) that is fixed by \(f\) is 0, which doesn't lie in \(S^2\).

We hope \(f\) turns out to be a homeomorphism.

\(||f(v)-f(w)||=||-v+w||=||v-w||\). So f is in fact Lipshitz function, so continuous.

\(f(f(v)=v\) for all \(v\in\mathbb{R}^3\). Therefore, \(f\) itself is inverse of \(f\). Which proves that \(f\) is bijective (since, inverse function exists) and homeomorphism (inverse is also continuous).

 

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TIFR 2013 problem 20 | Sequence limit~Fixed Point

Try this problem 20 from TIFR 2013 based on Sequence Limit - Fixed Point. This problem is an application of the Banach Fixed Point Theorem.

Question: TIFR 2013 problem 20

True/False?

Consider the function \(f(x)=ax+b\) with \(a,b\in\mathbb{R}\). Then the iteration \(x_{n+1}=f(x_n)\); \(n\ge0\) for a given \(x_0\) converges to \(\frac{b}{1-a}\) whenever \(0<a<1\).

Hint:

Banach Fixed Point Theorem

Discussion:

Once the existence of limit is guaranteed, we can safely calculate the limit. If limit is \(x\) then \( x=ax+b\). In other words, \(x=\frac{b}{1-a}\).

The question really is does limit exists?

\(x_{n+1}-x_{n}=f(x_{n})-f(x_{n-1}) \)

\(=ax_{n}-ax_{n-1}=a^{2}(x_{n-1}-x_{n-2})=...=a^{n}(x_{1}-x_{0})\).

For \(n<m\),

\(x_m-x_n=x_m-x_{m-1}+x_{m-1}-x_{m-2}+...-x_{n} \)

\( =(a^{m}+...a^{n})(x_1-x_0)=\frac{a^n}{1-a}(x_1-x_0) \)

And the right hand side converges to 0 as n tends to infinity (Here,we are using the fact that \(0<a<1\)) i.e, the right hand side can be made arbitrarily small using large enough n, so \(x_n\) is a Cauchy sequence. We are in \(\mathbb{R}\), so by completeness, \(x_n\) converges.

Remark:

The above discussion really didn't make use of Banach Fixed Point Theorem. But knowledge is power. And if known that :

"Let \(T:X\to X\) be a contraction mapping, X-complete metric space. Then T has precisely one fixed point \(u\in X\). Furthermore, for any \(x\in X\) the sequence \(T^k (x)\) converges and the limit is \(u\). " (Banach Fixed Point Theorem).

the answer is immediate. The above solution runs in the line of the proof of this Theorem.

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TIFR 2013 problem 19 | Non-uniformly continuous function

This problem from TIFR 2013, Problem 19 discusses the example of a non-uniformly continuous function.

Question: TIFR 2013 problem 19

True/False?

Every differentiable function $f:(0,1) \to [0,1]$ is uniformly continuous.

Hint: 

$\sin(1/x)$ is not uniformly continuous. However, range is not $[0,1]$. But its a simple matter of scaling.

Discussion:

Let $f(x)=\sin(\frac{1}{x})$ for all $x\in(0,1)$. For simplicity, we first prove that this $f$ is not uniformly continuous. Then we will scale things down, which won't change the non-uniform continuity of $f$.

Note that $f$ is differentiable. To show that $f$ is not uniformly continuous, we first note that as $x$ approaches $0$, $\frac{1}{x}$ goes through an odd multiple of  $\frac{\pi}{2}$ to an even multiple of  $\frac{\pi}{2}$ real fast. So in a very small interval close to $0$, I can find two such points which gives value $1$ and $0$.

Let $x=\frac{2}{(2n+1)\pi}$ and $y=\frac{2}{2n\pi}$.

Then $|x-y|=\frac{2}{2n(2n+1)\pi}$.

Since the right hand side of above goes to zero as $n$ increases, given any $\delta > 0$ we can find $n$ large enough so that $|x-y|<\delta$. For these $x$ and $y$, $|f(x)-f(y)|=1$.

Of course, choosing $\epsilon$ as any positive number less than $1$ shows that $f$ is not uniformly continuous.

We have with us a function $f$ which is bounded, differentiable and not uniformly continuous.

To match with the questions requirements, notice $-1\le f(x)\le 1$.

So $0\le 1+f(x) \le 2$ And $0\le \frac{1+f(x)}{2} \le 1$.

Define $g(x)= \frac{1+f(x)}{2}$ for all $x\in(0,1)$.

Since sum of two uniformly continuous functions is uniformly continuous and a scalar multiple of uniformly continuous function is uniformly continuous, if  $g(x)$ was uniformly continuous, then $f(x)=2g(x)-1$ would also be uniformly continuous.

This proves that g is in fact not uniformly continuous. It is still differentiable, and range is $[0,1]$, which shows that the given statement is actually false.

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TIFR 2013 problem 18 | Problem based on Limit

Try this problem from TIFR 2013 problem 18 based on limit.

Question: TIFR 2013 problem 18

True/False?

Let \(P(x)= 1+x+ \frac{x^2}{2!} +... \frac{x^n}{n!} \) where n is a large positive integer. Then \(\lim_{x\to\infty} \frac{e^x}{P(x)} = 1 \)

Hint:

n really doesn't matter!

Discussion:

As \(x\) approaches 'infinity', both \(e^x\) and \(P(x)\) tends to 'infinity' (i.e, gets arbitrarily large).

One could use L'Hospitals rule here. Without computation, the numerator \(e^x\) when differentiated gives \(e^x\) again. The denominator \(P(x)\) when differentiated will give a polynomial of degree \(n-1\). If \(n-1=0\), i.e, the polynomial obtained by differentiating is a constant one, then the limit is simply limit of \(e^x\) which is \(\infty\). If \(n-1 \neq 0 \) then again L'Hospital rule is applicable since any non-constant polynomial has limit \(\infty\).

Repeating the process sufficiently many times we will obtain a stage where numerator as always will remain \(e^x\) and denominator will be some constant polynomial. So the limit will be \(\infty\).

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TIFR 2013 Problem 17 Solution -Convergence of Improper Integral


TIFR 2013 Problem 17 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:True/False

True/ False?

The integral \(\int_{0}^{\infty} e^{-x^5}dx \) is convergent.

Hint:

As x varies from 0 to "infinity", \(-x^5\) varies from 1 to "minus infinity". Basically, the function is "rapidly decreasing". We can hope that nothing goes wrong in the convergence.

Discussion:

Recall that \(\int_{0}^{\infty} e^{-x}dx \) is convergent. (In fact the value of this integral is 1 which can be done by simple calculation). We try to use this for our comparison test. For \(1\le x < \infty \)   \(e^{-x^5} \le e^{-x} \). This is because in this interval, \(x^5 \ge x\) and because \(e^x\) is an increasing function. So for \(1\le x < \infty \), our worries are over.

\(\int_{1}^{\infty} e^{-x^5} \le \int_{1}^{\infty} e^{-x}dx < \infty \).

Now we only need to check whether the integration is finite on \([0,1]\). The intergand is continuous and bounded by 1 (because it is monotonic decreasing and value at 0 is 1) on this interval, hence the integral is bounded. \(\int_{0}^{1} e^{-x^5}dx \le \int_{0}^{1} 1dx = 1 \).

Thus, \(\int_{0}^{\infty} e^{-x^5}dx \) is convergent.

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TIFR 2013 Problem 16 Solution -Bounded or Not?


TIFR 2013 Problem 16 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:True/False

Suppose \(\left \{a_i\right\}\) is a sequence in \(\mathbb{R}\) such that \(\sum|a_i||x_i|<\infty\) whenever \(\sum|x_i|<\infty\). Then \(\left \{a_i\right\}\) is a bounded sequence.

Hint:

For any \(r\in(0,1)\), \(\sum r^n <\infty \).

Also, if the radius of convergence of a power series is R, then R is  given by \(limsup|a_n|^{1/n}=\frac{1}{R} \)

Discussion:

Of course, \(\sum|a_n||r|^{n}<\infty\) for any \(r\in(-1,1)\).

Recall that, \(\sum|a_n|x^{n}<\infty\) for \(|x|<m\) means that radius of convergence of the power series is atleast m.

If the radius of convergence of  \(\sum|a_n|x^{n}\) is R then \(R\ge1\).

i.e, $$  limsup|a_n|^{1/n} = \frac{1}{R} \le 1  $$

Hence, there exists \(N\in \mathbb{N}\) such that \(sup\{|a_n|^{1/n} : n \ge k\} \le 1 \) for all \(k \ge N\) (This is from the definition of limsup of a sequence).

Hence,  \(sup\{|a_n|^{1/n} : n \ge N\} \le 1 \). Therefore, for each \(n \ge N\) we have \(|a_n|^{1/n} \le 1 \) for all \(n \ge N\). (Because sup is supremum which is least upper bound).

A real number which is in between 0 and 1 when raised to any power stays in between 0 and 1.

This allows us to state that \(|a_n| \le 1\) for all \(n \ge N\).

There are only finitely many terms left in the sequence which may not bounded by 1. But taking the maximum of their absolute value and 1 together we get a bound for the whole sequence.

For any \(n\in \mathbb{N}\),

$$ |a_n| \le max\{1,|a_1|,|a_2|,...,|a_{N-1}| \} $$

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TIFR 2013 problem 15 | Problem on Convergence

Try this problem based on convergence from TIFR 2013, Problem 15.

Question: TIFR 2013 problem 15

True/False?

Let \(x_1\in(0,1)\). For \(n>1\) define \(x_{n+1}=x_n-x_n^{n+1} \) Then \(lim_{n\to \infty} x_n \) exists.

Hint: A bounded monotone sequence is convergent.

Discussion:

By induction, we can prove that the sequence is between 0 and 1 always.

Suppose, \(x_n\in(0,1)\). Since we are removing a positive number from \(x_n\) to get \(x_{n+1}\), we have \(x_{n+1}<x_n<1\). This also shows that the sequence is decreasing. And since for a number in between 0 and 1, when we take positive powers it decreases, in other words \(x_n>x_n^{n+1}\) we have \(x_{n+1}>0\).

Therefore the given sequence is decreasing and bounded below by 0. Hence it is convergent.

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