Continuous Image of Connected Space is Connected
Theorem
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 $f \sqbrk {S_1}$ 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 $f \sqbrk T$ is a real interval.
Corollary 3
Let $I = \closedint a b$ be a closed real interval.
Let $f: I \to \R$ 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 $T_1$.
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$\blacksquare$
Proof 2
Suppose that $S_2 = f \sqbrk {S_1}$ is not connected in $T_2$.
Then by definition there exist open sets $U_2$ and $V_2$ in $T_2$ such that:
- $S_2 \subseteq U_2 \cup V_2$
- $U_2 \cap V_2 \cap S_2 = \O$
- $U_2 \cap S_2 \ne \O$
- $V_2 \cap S_2 \ne \O$
We have by hypothesis that $f: T_1 \to T_2$ is continuous.
Thus $U_1 = f^{-1} \sqbrk {U_2}$ and $V_1 = f^{-1} \sqbrk {V_2}$ are open in $T_1$.
We have that:
- $U_2 \cap S_2 \ne \O$
Therefore:
- $\exists x \in S_1: \map f x \in U_2$
Then:
- $x \in f^{-1} \sqbrk {U_2} = U_1$
and:
- $x \in S_1$
so:
- $U_1 \cap S_1 \ne \O$
Similarly:
- $V_1 \cap S_1 \ne \O$
Suppose there exists $x \in S_1$ such that $x \in U_1 \cap V_1 \cap S_1$.
Then:
- $\map f x \in U_2 \cap V_2 \cap S_2$
which is a contradiction.
It follows that:
- $U_1 \cap V_1 \cap S_1 = \O$
Thus by definition $S_1$ is not connected in $T_1$.
The result follows by the Rule of Transposition.
$\blacksquare$
Sources
- 1953: Walter Rudin: Principles of Mathematical Analysis ... (next): $4.22$