Continuous Image of Connected Space is Connected

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Let $T_1$ and $T_2$ be topological spaces.

Let $S_1 \subseteq T_1$ be connected.

Let $f: T_1 \to T_2$ be a continuous mapping.

Then the image $f \left({S_1}\right)$ is connected.

Corollary 1

Connectedness is a topological property.

Corollary 2

Let $T$ be a connected topological space.

Let $f: T \to \R$ be a continuous real-valued mapping.

Then $f \left({T}\right)$ is a real interval.

Corollary 3

Let $\mathbb I = \left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $f: \mathbb I \to \R$ be a continuous mapping.

Then $f$ has the intermediate value property.

Proof 1

Let $i: f \sqbrk {T_1} \to T_2$ be the inclusion mapping.

Let $g: T_1 \to f \sqbrk {T_1}$ be the surjective restriction of $f$.

Then $f = i \circ g$.

Hence, by Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is continuous.

We use a Proof by Contradiction.

Suppose that $A \mid B$ is a partition of $f \sqbrk {T_1}$.

Then it follows that $g^{-1} \sqbrk A \mid g^{-1} \sqbrk B$ is a partition of $T_1$.


Proof 2

Suppose that $S_2 = f \left({S_1}\right)$ is not connected in $T_2$.

Then by definition there exist open sets $U_2$ and $V_2$ in $T_2$ such that:

$S_2 \subseteq U_2 \cup V_2$
$U_2 \cap V_2 \cap S_2 = \varnothing$
$U_2 \cap S_2 \ne \varnothing$
$V_2 \cap S_2 \ne \varnothing$

By hypothesis, $f: T_1 \to T_2$ is continuous.

Thus $U_1 = f^{-1} \left({U_2}\right)$ and $V_1 = f^{-1} \left({V_2}\right)$ are open in $T_1$.

We have that:

$U_2 \cap S_2 \ne \varnothing$


$\exists x \in S_1: f \left({x}\right) \in U_2$


$x \in f^{-1} \left({U_2}\right) = U_1$


$x \in S_1$


$U_1 \cap S_1 \ne \varnothing$


$V_1 \cap S_1 \ne \varnothing$

Suppose there exists $x \in S_1$ such that $x \in U_1 \cap V_1 \cap S_1$.


$f \left({x}\right) \in U_2 \cap V_2 \cap S_2$

which is a contradiction.

It follows that:

$U_1 \cap V_1 \cap S_1 = \varnothing$

Thus by definition $S_1$ is not connected in $T_1$.

The result follows by the Rule of Transposition.