User:Caliburn/s/8

Theorem

Let $\struct {X, \tau_1}$ be a connected topological space.

Let $\struct {Y, \tau_2}$ be a topological space.

Let $f : X \to Y$ be a locally constant function.

Then $f$ is constant.

Proof

Suppose that there exists some locally constant $f$ that is not constant.

That is, there at least two distinct points in $\map f X$.

We show that $X$ is then disconnected.

Let $y \in \map f X$.

Since $f$ is locally constant, for each $x \in X$ such that:

$y = \map f x$

there exists an open set $U_x$ such that:

$\map f {U_x} = \set y$

We now prove that:

$\ds \map {f^{-1} } {\set y} = \bigcup_{x : y = \map f x} U_x$

For any $x \in \map {f^{-1} } {\set y}$, we have $\map f x = y$, so:

$x \in U_x$

and hence:

$\ds x \in \bigcup_{x : y = \map f x} U_x$

giving:

$\ds \map {f^{-1} } {\set y} \subseteq \bigcup_{x : y = \map f x} U_x$

Now let:

$\ds u \in \bigcup_{x : y = \map f x} U_x$

then:

$u \in U_x$

for some $x$ with $y = \map f x$.

Since:

$\map f {U_x} = \set y$

we then have:

$\map f u = y$

so:

$u \in \map {f^{-1} } {\set y}$

giving:

$\ds \map {f^{-1} } {\set y} = \bigcup_{x : y = \map f x} U_x$

From the definition of a topological space, we have that:

the union of open sets is open.

So:

$\map {f^{-1} } {\set y}$ is open for each $y \in \map f X$.

We therefore have:

\(\ds \bigcup_{y \in \map f X} \map {f^{-1} } {\set y}\)

\(=\)

\(\ds \map {f^{-1} } {\bigcup_{y \in \map f X} \set y}\)

Preimage of Union under Mapping: Family of Sets

\(\ds \)

\(=\)

\(\ds \map {f^{-1} } {\map f X}\)

\(\ds \)

\(=\)

\(\ds X\)

Preimage of Mapping equals Domain

Fix some $y_* \in \map f X$.

Since $\map f X$ contains at least two distinct points:

$\map f X \setminus \set {y_*} \ne \O$

We then have:

$\ds X = \map {f^{-1} } {\set {y_*} } \cup \bigcup_{y \in \map f X \setminus \set {y_*} } \map {f^{-1} } {\set y}$