3.2 Connectedness

Proof. Observe any disjoint open (resp. closed) covering of a topological space by 2 sets is also a closed (resp. open) covering hence (C1) is equivalent to (C2)

Suppose (C2) holds and $f:X\to \curl{0,1}$ is surjective then $\curl{f\inv(\curl{0}),f\inv(\curl{1})}$ is disjoint cover of $X$ by 2 sets so $f\nin C(X;\curl{0,1})$ which shows (C3).

Suppose (C3) holds. Let $U\suf X$ is clopen then the characteristic function

is continuous hence constant showing (C4).

Suppose (C4) holds. If $\curl{U_i}_{i\in I}$ is a disjoint open cover of $X$ by nonempty sets then fix $j\in I$. Observe $X\del U_{j}=\cups{}{j\neq i\in I }U_{i},U_{j}$ are open so by (C4) $U_{j}=X$ then for any $i\in I$ we have $U_{i},U_{j}$ is the desired pair of nondisjoint sets.

By hypothesis (C5) implies (C1).$\square$

Proof. Let $\curl{U,V}\suf \cali{P}(Y)$ be an open cover of $f(X)$ in the sense that $f(X)\suf U\cap V$. By continuity of $f$ and connectedness of $X$ we have $f\inv(U)\cap f\inv(V)=f\inv(U\cap V)\neq\emp\implies U\cap V\neq\emp$.$\square$

Proof. Let $f\in C(A;\curl{0,1})$ then by Definition 3.1.10

1. (1)

   Observe $f|_{Z}$ is continuous hence $f\equiv c\in \curl{0,1}$ so $f\inv(c)$ is closed and contains $Z$. We conclude $A=\cl{Z}\suf f\inv(c)$ showing $f$ is constant.

2. (2)

   Let $p\in \caps{}{i\in I }A_{i}$ then for each $i\in I$, $f|_{A_i}\equiv f(p)$ hence $f(A)=f(p)$.

$\square$

Proof. By Proposition 3.2.3(2) this is an equivalence relation and since the connected component containing $X$ is the largest connected subset of $X$ containing $x$ it is closed by Proposition 3.2.3(1).$\square$