# Connectedness and continuity

• Nov 5th 2013, 07:12 PM
mrmaaza123
Connectedness and continuity
The following is a theorem from my book, but i don't think that i follow the proof given:

Theorem:
Every continuous image of a connected metric space is connected.

Proof:
Suppose X and Y are metric spaces and g : X -> Y is continuous.

Suppose g(X) is not a connected space.

Then there exists a continuous function f from g(X) onto {0,1}. (i don't think i follow this)

This yields (fog)(X) = f(g(X)) = {0,1}, and since fog is continuous,

X does not satisfy the continuity criterion for connectedness and therefore is not connected. (I don't understand how this happens ? why is it not satisfying the continuity criterion ?)

Any help will be appreciated.
• Nov 5th 2013, 07:22 PM
SlipEternal
Re: Connectedness and continuity
Suppose $g(X)$ is not connected. Then, $g(X) = U \cup V$ where $U\cap V = \emptyset$, and both $U$ and $V$ are open and nonempty. Define $f:g(X) \to \{0,1\}$ by $f(x) = \begin{cases}0 & \text{if }x \in U \\ 1 & \text{if }x \in V\end{cases}$. Now, $f$ is continuous if the preimage of open sets are open. So, let's look at the open sets of $\{0,1\}$. They are $\{\emptyset, \{0\}, \{1\}, \{0,1\}\}$. The preimage of the empty set is the empty set, which is open in $g(X)$. The preimage of $\{0\}$ is $U$, which is open in $g(X)$. The preimage of $\{1\}$ is $V$, which is open in $g(X)$, and the preimage of $\{0,1\}$ is $g(X)$, which is obviously open in $g(X)$. So, $f$ is continuous. You have already seen the proof that the composition of continuous functions is continuous. So, $(f\circ g):X \to \{0,1\}$ must be a continuous function. Hence, $(f\circ g)^{-1}(\{0\}), (f\circ g)^{-1}(\{1\})$ must be disjoint, nonempty, open subsets of $X$ whose union is $X$. This implies $X$ is not connected.

Edit: Disjoint comes from the definition of a function. If they were not disjoint, then let $x \in (f\circ g)^{-1}(\{0\}) \cap (f\circ g)^{-1}(\{1\})$. Then $(f\circ g)(x) = 0$ and $(f\circ g)(x) = 1$ implying $(f\circ g)$ is not even a function.

Nonempty comes from the fact that $U,V$ are nonempty subsets of $g(X)$, so $g^{-1}(U)$ and $g^{-1}(V)$ must be nonempty subsets of $X$.

Open comes from continuity of $(f\circ g)$.

The union $(f\circ g)^{-1}(\{0\}) \cup (f\circ g)^{-1}(\{1\}) = (f\circ g)^{-1}(\{0,1\}) = X$ because the preimage of the entire image of a function is the entire domain.