Theorem about connectedness.

Question

$(X, \mathcal{O})$ : topological space

$\{ M_{\lambda} | \lambda \in \Lambda \}$ : subset system of X

Theorem;

\begin{equation} \forall \lambda \in \Lambda ; M_{\lambda} \text{ is connected space}, x \in M_{\lambda} \Longrightarrow M:=\cup_{\lambda \in \Lambda} M_{\lambda} \text{ is connected space.} \end{equation}

I could understand the proof partially, but I couldn't understand the last part.

Proof;

Let $N\subset M$ is open and closed in $(M, \mathcal{O_M})$.

$\exists \, G\text{(open in } (X,\mathcal{O})), F\text{(closed in }(X,\mathcal{O})); N= M\cap G=M\cap F.$

Then, $N\cap M_{\lambda} =M_{\lambda}\cap G=M_{\lambda}\cap F.$

Thus, $N\cap M_{\lambda}$ is open and closed in $(M_{\lambda}, \mathcal{O_{M_{\lambda}}})$

Since $(M_{\lambda},\mathcal{O_{M_{\lambda}}} )$ is connected space, $N\cap M_{\lambda}= M_{\lambda} \text{ or } \emptyset.$

And I have to prove $N=M \text{ or } \emptyset.$

I could prove that $\forall \lambda \in \Lambda; N\cap M_{\lambda} =\phi \Longrightarrow N=\emptyset.$

But I cannot prove that $\exists \lambda \in \Lambda; N\cap M_{\lambda}=M_{\lambda} \Longrightarrow N=M.$

I would like you to give me ideas about why this holds.

Answer 1

You missed the crucial part of the hypothesis. It is given that there is a point $x$ which belings to every $M_{\lambda}$. Suppose $x \in N$. Then $x \in N \cap M_{\lambda}$ so the first option does not arise and you get $N \cap M_{\lambda}= M_{\lambda}$ for every $\lambda$ which gives $N=M$.

Now suppose $x \notin N$. Then the second option does not arise so we get $N \cap M_{\lambda}=\emptyset $ for every $\lambda$. This gives $N=\emptyset$.

Answer 2

I'll assume (as in common in this result) that $x$ is some fixed point of $X$, common to all $M_\lambda$.

Suppose $C$ is a clopen and non-empty subset of $M$. We can assume WLOG that $x \in C$ (or we take $M\setminus C$ instead, which is then also non-empty clopen in $M$ and then contains $x$).

Fix $\lambda \in \Lambda$. Then $C \cap M_\lambda$ is clopen in $M_\lambda$ (as each $M_\lambda$ has the subspace topology w.r.t. $M$). It still contains $x$ so it's non-empty. As $M_\lambda$ is connected, $M_\lambda \cap C = M_\lambda$, or equivalently $M_\lambda \subseteq C$.

As $\lambda$ was arbitrary, $M \subseteq C$ and so $C=M$ and we're done: $M$ is connected.