Let
be
the unit interval, with the Subspace Topology. Let
also have the Subspace Topology.
There does not exist a continuous function
such that
and
.
Proof:
Let
. Let
Since
we know
We
also know that
for
Suppose
Then
since
is continuous there exists a
such
that
for all
contradiction the
previous statement.
Hence
.
Therefore
there exists a
such
that
for all
But since
for
any
there exists an
such that
.
In particular
But
,
again a contradiction.
Proving that such an
does not exist.
There does not exist a continuous, onto, function
.
Proof:
Choose
such
that
and
.
Define
by the formula
Then
would be a continuous function
such that
and
Such
functions were shown not to exist in the Theorem above.
Let
be
the unit
disk,
. Let
be the unit circle.
There does not exist a continuous function
such that
is the identity map,
for
1. Let
be
the unit disk, . Let
be the unit sphere.
There does not exist a continuous function
such that
is the identity map.
2. Every polynomial with complex coefficients has a complex root.