Let $X\subset M$ is na open and closed set such as $x \in X, y \in X^c$. Then, there is no connected in M wich contains x and y.

Question

Let $X\subset M$ is an open and closed set such as $x \in X, y \in X^c$. Proof that there is no connected in M which contains x and y.

What I did:

Suposse that there is $Y \subset M$ connected with $x,y \in Y$. So $Y \cap X \neq \emptyset $ and $Y \cap X^c \neq \emptyset$, then $Y \cap \partial X \neq \emptyset$

How can I argue that there is a contradiction?

I want to show that there is no Y connected in M with $x, y \in Y$ — user 242964, Sep 20 '15 at 17:00

score 1 · Accepted Answer · answered Sep 20 '15 at 17:01

Suppose there is a connected set $Y$ s.t $ x,y \in Y$ then consider $X \cap Y$ and $X^c \cap Y$. Now $X$ is both open and closed. So, $X \cap Y$ and $X^c \cap Y$ are both open in $Y$ and their intersection is $\phi$. That means $Y$ is disconnected. Contradiction.

Let $X\subset M$ is na open and closed set such as $x \in X, y \in X^c$. Then, there is no connected in M wich contains x and y.

1 Answers1