Please is this prof is correct ?
Let $\{\Omega_i\}_{i\in I}$ a famille of connected sets such that $$\forall i,j\in I, \Omega_i\cap\Omega_j\neq\emptyset$$
I want to prove that $\bigcup_{i\in I} \Omega_i$ is connected.
If I suppose that $\bigcup \Omega_i$ is not connected then, there exists two non empty open sets $A,B$ from $\bigcup \Omega_i$ such that
$$
\begin{cases}
\bigcup \Omega_i= A\cup B\\
A\cap B=\emptyset
\end{cases}
$$
we have $\forall i\in I, \Omega_i\subset \bigcup_{i\in I}\Omega_i=A\cup B$ then by the connectedness of $\Omega_i$
$$\forall i\in I, [\Omega_i\subset A ~\text{or}~ \Omega_i\subset B]$$
As $\forall i,j\in I, A_i\cap A_j\neq \emptyset$ we deduce that $$\forall I\in A, \Omega_i\subset A~\text{or}~ \forall i\in I,\Omega_i\subset B$$
it follows that $B=\emptyset$ or $D=\emptyset$, which is a contradiction.
Please if I change the condition
$$\forall i,j\in I, \Omega_i\cap\Omega_j\neq\emptyset$$
by
$$
\exists i_0\in I, \Omega_{i_0}\cap \Omega_j\neq \emptyset,\forall j\in I
$$
How to do ?
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Let $\{\Omega_i\}_{i\in I}$ a famille of connected sets such that $$\exists i_0\in I, \Omega_{i_0}\cap \Omega_j\neq \emptyset,\forall j\in I$$
I want to prove that $\bigcup_{i\in I} \Omega_i$ is connected.
If I suppose that $\bigcup \Omega_i$ is not connected then, there exists two non empty open sets $A,B$ from $\bigcup \Omega_i$ such that
$$
\begin{cases}
\bigcup \Omega_i= A\cup B\\
A\cap B=\emptyset
\end{cases}
$$
we have $\forall i\in I, \Omega_i\subset \bigcup_{i\in I}\Omega_i=A\cup B$ then $\Omega_{i_0}\subset A\cup B$ as it is connected we have $$\Omega_{i_0}\subset A~\text{or} ~ \Omega_{i_0}\subset B$$ if we suppose that $\Omega_{i_0}\subset A$ then $$\forall j\in I, \Omega_{j}\cap A\neq \emptyset $$ by the connectedness of $\Omega_i$ we deduce that $$\forall j\in I, \Omega_{j}\cap B=\emptyset$$ then $$\forall j\in I, \Omega_j\subset A$$ thus $$\bigcup_{j\in I}\Omega_j\subset A$$ so $B=\emptyset$ contradiction in the same way if we suppose that $\Omega_{i_0}\subset B$ we find that $A=\emptyset$
thank you
Best Answer
Actually, there is a much easier way to prove any given set is connected. Think of it as the "magic wand" for proving connectedness of sets.
Suppose $X$ is a set. Then, it is connected iff every continuous function $f: X \rightarrow \left\lbrace 1, -1 \right\rbrace$ is constant.
Your second case can be easily proven by this fact. Notice that each $\Omega_i$ is connected and hence every continuous function $f_i: \Omega_i \rightarrow \left\lbrace 1, -1 \right\rbrace$ is constant. Now, this is true also for $\Omega_{i_0}$. Now, also observe that $\forall i \in I$, $\Omega_{i_0} \cap \Omega_i \neq \emptyset$ implies that $\exists x_0 \in \Omega_i$ such that it is also in $\Omega_{i_0}$. Now, since $f_{i_0}$ is a constant function, its value at $x_0$ is constant and must be same everywhere in $\Omega_{i_0}$ as well as $\Omega_i$ for every $i \in I$. This is because each $\Omega_i$ is connected. Hence, every continuous function $f : \bigcup\limits_{i \in I} \Omega_i \rightarrow \left\lbrace 1, -1 \right\rbrace$ is constant and hence $\bigcup\limits_{i \in I} \Omega_i$ is connected.