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$(X, \mathcal{O})$ : 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.

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    $\begingroup$ The idea is that N has either to be the empty set or M. You cannot prove both, but you can only prove that one of either holds. So, I'd be interested to see your proof for N = emptyset. I think you are already pretty close to solving the problem $\endgroup$ – don-joe Jan 26 at 8:30
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    $\begingroup$ Note that you're not using the common point $x$ at all. You need it. $\endgroup$ – Henno Brandsma Jan 26 at 8:32
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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 $N=\emptyset$.

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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.

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