A Fundamental Theorem

Theorem:

Let MATH MATHbe the unit interval, with the Subspace Topology. Let MATH also have the Subspace Topology.

There does not exist a continuous function MATH such that MATHand MATH.

Proof:

Let MATH . Let MATHSince MATH we know $\QTR{Large}{l<1.}$We also know that MATH for $\QTR{Large}{x>l.}$

Suppose MATH

Then since $\QTR{Large}{f}$ is continuous there exists a MATHsuch that MATH for all MATH contradiction the

previous statement.

Hence MATH .

Therefore there exists a MATHsuch that MATH for all MATH

But since MATH for any MATH there exists an MATH such that

MATH. In particular MATH But MATH, again a contradiction.

Proving that such an $\QTR{Large}{f}$ does not exist.

Corollary:

There does not exist a continuous, onto, function MATH .

Proof:

Choose MATHsuch that MATHand MATH. Define MATH by the formula

MATH Then MATH would be a continuous function MATH such that MATHand MATHSuch functions were shown not to exist in the Theorem above.

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Later in the Semester:

Let MATHbe the unit disk,MATH . Let MATH be the unit circle.

There does not exist a continuous function MATH such that MATH is the identity map, MATH for MATH

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Coming Attractions:

1. Let MATHbe the unit disk, . Let MATH be the unit sphere.

There does not exist a continuous function MATH such that MATH is the identity map.

2. Every polynomial with complex coefficients has a complex root.