Topology Proof Advice/Guidance

Question

I have been working on my final proof in my Fundamentals of Mathematics course. It is (supposedly) a basic topological proof... I cannot say for sure because the topic is so new to me. I have written up an extremely rough draft of it, as I am more focused right now on making sure I actually understand the basic ideas of connectedness and spaces as opposed to writing an excellent proof. The rough draft is due tomorrow and the final draft not until next week, so I do have some time. I was just looking for thoughts, ideas and other constructive criticisms that will help me understand these ideas better and help me to write this proof. The topic of topology is very new to the class so I don't believe the proof is meant to be very difficult... but the understanding of new ideas is supposed to get us thinking.

Proposition

If $E$ and $F$ are connected subsets of $M$ (where $E \cap F \neq \emptyset$), then $E \cup F$ is also connected.

Proof

Suppose it is not the case that $E \cup F$ is connected. Then $E \cup F$ is disconnected. By definition, a metric space $N$ is disconnected if $N = A \cup B$, where $A$ and $B$ are disjoint open sets in $N$. If $E \cup F$ is disconnected, this implies that there are no elements appearing in $E$ that also appear in $F$, and vice versa. Hence, we can say that $E \cap F = \emptyset$, which means that $E$ and $F$ cannot be connected subsets within a metric space $M$.

Therefore, by contraposition, if $E$ and $F$ are connected subsets of $M$ (where $E \cap F \neq \emptyset$), then $E \cup F$ is also connected.

P.S. I was told by my professor that a proof by contrapositive would be a good way to go about this proof, as it is easier to prove disconnectedness than connectedness, so that is the reasoning behind that decision.

Answer 1

At some point you refer to metric spaces. This may be a typo of some sort. The result is true for all topological spaces.

One way to simplify problems such as this is to use the "transtivity" of sub-space topologies: If $A\subset B\subset M$ and $B$ has the sub-space topology "inherited" from $M.$ then the topology on $A$ as a sub-space of $B$ is the same as the topology on $A$ as a sub-space of $M.$

So since $E$ and $F$ are connected sub-spaces of $M$, they are connected sub-spaces of $E\cup F.$

By contradiction, suppose $E\cup F$ is disconnected. Then $E\cup F=A_0\cup A_1$ where $A_0,A_1$ are disjoint non-empty open subsets of the space $E\cup F.$

Then $E_0=E\cap A_0$ and $E_1=E\cap A_1$ are disjoint open subsets of $E,$ and their union is $E.$ But $E$ is connected, so one of $E_0, E_1$ is empty and the other is $E.$

Similarly, with $F_0=F\cap A_0$ and $F_1=F\cap A_1$, one of $F_0,F_1$ is empty and the other is $F.$

Suppose $E_0=E$ and $E_1=\phi.$ Then $$\phi\ne E\cap F=E_0\cap\ F=(E\cap A_0)\cap F \subset A_0\cap F=F_0.$$ So $F_0=F$ and $F_1=\phi$. But then $$A_1=A_1\cap (E\cup F)=(A_1\cap E)\cup (A_1\cap F)=E_1\cup F_1=\phi,$$ contrary to $A_1\ne \phi.$

Interchanging the subscripts $0,1$ in the above paragraph, we also obtain a contradiction if $E_1=E$ and $E_0=\phi.$

The valid generalization is that if $S$ is any non-empty family of connected sub-spaces of $M,$ and $\cap S\ne \phi,$ then $\cup S$ is connected.

Answer 2

If $E\cup F$ is disconnected, this implies that there are no elements appearing in $E$ that also appear in $F$...

It is true (assuming all the other hypothesis on $E,F$), because it is what you need to prove. But why is it true? Definition of non-connectedness just says that there are two disjoint sets $A$, $B$ such that $A\cup B=E\cup F$ and $A,B$ are open in $E\cup F$. Of course, even if you had the hypothesis that $E$ and $F$ are open in $E\cup F$ (and you don't), this needs not imply that $E=A$ and $F=B$.

I'd rather suggest assuming, without loss of generality, that $E\cap A\ne\emptyset$ and then start speculating whether or not $E\subseteq A$ must be true. After that, draw similar conclusions for $F$ by looking at $E\cap F$.