# Thread: Path connectedness and connectedness

1. ## Path connectedness and connectedness

Problem

Show that an open set $\Omega \in \field{C}$ is pathwise connected if and only if it is connected.

Attempt at Solution

Suppose $\Omega$ is not connected and suppose $\Omega$ is pathwise connected. Then, for any points $a,b \in \Omega$ there exists a continuous function $f$ which connects $a$ and $b$ together. Let $V,W \subset \Omega$, where $V \cap W = \emptyset$ and $\Omega = V \cup W$. Let $a \in V$ and $b \in W$. Clearly, we have a contradiction because there cannot exist a continuous function from $a$ to $b$ if $\Omega$ is disconnected. Thus $\Omega$ must be connected.

How would I prove the other direction? Is this correct, btw?

2. ## Re: Path connectedness and connectedness

Originally Posted by My Little Pony
Problem

Show that an open set $\Omega \in \field{C}$ is pathwise connected if and only if it is connected.

Attempt at Solution

Suppose $\Omega$ is not connected and suppose $\Omega$ is pathwise connected. Then, for any points $a,b \in \Omega$ there exists a continuous function $f$ which connects $a$ and $b$ together. Let $V,W \subset \Omega$, where $V \cap W = \emptyset$ and $\Omega = V \cup W$. Let $a \in V$ and $b \in W$. Clearly, we have a contradiction because there cannot exist a continuous function from $a$ to $b$ if $\Omega$ is disconnected. Thus $\Omega$ must be connected.

How would I prove the other direction? Is this correct, btw?
More generally, if $V$ is a normed vector space then any connected open subset $U$ of $V$ is path connected.

This follows from the fact that ever locally path connected and connected space is path connected. But, we know that $U$ is connected and since every point of $U$ has some open ball $B$ containing it sitting inside $U$, but since $B$ is convex and thus path connected you have that $U$ is locally path connected.

If you want to learn more, or see proofs of the above you can look at my blog post here (sorry, the formatting on that one is pretty bad).