TIFR 2014 Problem 29 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 continuous map between metric spaces. Then
is a complete subset of
if
A. X is compact
B. Y is compact
C. X is complete
D. Y is complete
Discussion:
Let be compact. Then
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 be a Cauchy sequence in
. Then since
is compact,
has a convergent subsequence (converging to a point in that compact set i.e, in
.
Suppose .
Then by triangle inequality, we have as
Here we have used that is cauchy to conclude
.
So this implies . Since
we conclude that
is complete.
This proves A.
Let and
. Take the inclusion map
for all
. This example shows that even if we take
to be compact, or complete,
need not be complete. So this disproves B and D.
Now take and take
. 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.
TIFR 2014 Problem 29 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 continuous map between metric spaces. Then
is a complete subset of
if
A. X is compact
B. Y is compact
C. X is complete
D. Y is complete
Discussion:
Let be compact. Then
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 be a Cauchy sequence in
. Then since
is compact,
has a convergent subsequence (converging to a point in that compact set i.e, in
.
Suppose .
Then by triangle inequality, we have as
Here we have used that is cauchy to conclude
.
So this implies . Since
we conclude that
is complete.
This proves A.
Let and
. Take the inclusion map
for all
. This example shows that even if we take
to be compact, or complete,
need not be complete. So this disproves B and D.
Now take and take
. 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.