# 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): $4.22$