Q1
Let
be an open set in
Since
such that
are continuous functions, therefore
is open in
and
and
is open in
.
are open sets. Since
, therefore
, which is open in
. Therefore,
is
continuous.
Q2
(a)
Consider the map
defined as
for each
, which is a constant function that is continous. If
this is a quotient of two continous function,
argument, the map
continous.
defined as
in
. If
,
,
. Since
is continous as well. By the similar
for each
in
,
is
(b)
if
. However,
if
.
(c)
Consider
, which is not closed. Therefore,
is not
continous.
Q3
Consider
defined as
. The inverse function
is
. Consider an open set in
,
be
, which is open in
. Consider an open set
,
, which is open in
. Both prove
and its inverse function is continuous, which implies that
is homeomorphic with
Q4
=>
in
.
Consider
be an arbitrary topological space in
that
. Let
be an open set in
. Since the subbasis elements are open set,
. Given that the sequence
that
for all
such
is open set, and
converges to , therefore, exists
, which implies that
for
such
.
<=
Consider an open set
containing point
form of
denote
where
in
. Denote
in
will be the
except finite of them denote the open sets that are not
, and
.
is open in
converges to
such that
, there
.
such that
for
. Therefore, define
,
for
. Therefore,
Hence, for
proves
in product topology.
be the indices for these
for
exists
in
for
, which
converges to .
If we consider
be open in terms of the box topology, then
Consider sequence
sequence
in
where we define
. For each
will converges to point 0. However,
in box topology. Consider an open set
, the
doesn't converges to point
in box topology defined as
such that
then exists
is not true in general.
. Assume
such that
converges to
, which is
, which
is wrong.
Q5
It's trivial to see
topology.
is bijective. Consider an open set
will be the form of
open sets that are not
to be
in
where
in
denote
. Denote
equipped with product
except finite of them denote the
be the indices for these
. In this sense,
. Since
is a linear function, it's a
. Define
homeomorphism of
equal to
, and for
with itself. Therefore, for
,
,
equals to
is open but not
. Therefore,
is open in
equipped with product topology. By a similar argument, we can prove that
in
equipped with product topology. Therefore,
is a homeomorphism of
is o[en
with itself.