TIFR 2014 Problem 28 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
Let be a topological space such that every function
is continuous. Then
A. has the discrete topology.
B. has the indiscrete topology.
C. is compact.
D. is not connected.
Discussion:
We know that if is a discrete space then any function
is continuous.
Option A asks whether the converse to this is true in the case that .
To prove/disprove whether has the discrete topology or not it is enough to prove whether every singleton set is open or not.
If we can show that for every there exists a function
such that
then we are done. Because we are given that
if exists must be continuous, and since
is open in
we will have the inverse image of it open in
, so
will be open in
.
Now, this target is easy to handle. We define for each
and
for
.
This satisfies our desired property. So
is discrete.
Taking (for example) shows that
does not need to be indiscrete nor does it have to be compact.
Taking shows that
may be connected. Of course if
has cardinality more than 1, it is not connected.
TIFR 2014 Problem 28 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
Let be a topological space such that every function
is continuous. Then
A. has the discrete topology.
B. has the indiscrete topology.
C. is compact.
D. is not connected.
Discussion:
We know that if is a discrete space then any function
is continuous.
Option A asks whether the converse to this is true in the case that .
To prove/disprove whether has the discrete topology or not it is enough to prove whether every singleton set is open or not.
If we can show that for every there exists a function
such that
then we are done. Because we are given that
if exists must be continuous, and since
is open in
we will have the inverse image of it open in
, so
will be open in
.
Now, this target is easy to handle. We define for each
and
for
.
This satisfies our desired property. So
is discrete.
Taking (for example) shows that
does not need to be indiscrete nor does it have to be compact.
Taking shows that
may be connected. Of course if
has cardinality more than 1, it is not connected.