Continuous Image of Connected Space is Connected

From ProofWiki
Jump to navigation Jump to search

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 $T_1$.


This article, or a section of it, needs explaining.
In particular: Add the justification for the above - may already exist as a result, otherwise add a proof something like Preimage of Partition is Partition.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Explain}} from the code.


$\blacksquare$


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

Retrieved from "https://proofwiki.org/w/index.php?title=Continuous_Image_of_Connected_Space_is_Connected&oldid=585744"