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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 ? %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 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

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    $\begingroup$ Just try doing the same thing; it is very similar. $\endgroup$ – 4-ier Aug 23 '18 at 10:45
  • $\begingroup$ @4-ier how to prove with the new condition that $\forall I\in A, \Omega_i\subset A~\text{or}~ \forall i\in I,\Omega_i\subset B$ $\endgroup$ – Vrouvrou Aug 23 '18 at 11:17
  • $\begingroup$ @Vrouvrou Yes, your proof does seem correct. However, I would still recommend to use the "magic wand" stated below whenever you wish to prove/disprove connectedness. This is because not always is it feasible to find the two non - empty open sets $A$ and $B$ such that all the conditions hold. Moreover, such a definition can be used for (full) spaces but it is really really difficult to use this definition for subspaces. Indeed, we can always use it because all definitions are equivalent. $\endgroup$ – Aniruddha Deshmukh Aug 25 '18 at 4:58
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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.

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  • $\begingroup$ Is it the same thing if we suppose that $\bigcap_{I\in I} A_i\neq\emptyset$? $\endgroup$ – Vrouvrou Aug 23 '18 at 11:18
  • $\begingroup$ Not really. $\bigcap\limits_{i \in I} A_i \neq \emptyset$ implies that the overall intersection is non - empty, which means that all sets have at least one point in common. Here, it is mentioned that $\exists i_0 \in I$ such that $\forall i \in I$, $A_i \cap A_{i_0} \neq \emptyset$. This means that every set has at least one point common with the set $A_{i_0}$ but that does not mean all the sets have common points with one another. $\endgroup$ – Aniruddha Deshmukh Aug 23 '18 at 11:32
  • $\begingroup$ can you explain me how to prove with this new condition? and does my proof still correct ? $\endgroup$ – Vrouvrou Aug 23 '18 at 11:37
  • $\begingroup$ Go for the same way. If $\bigcap\limits_{i \in I} A_i \neq \emptyset$, then $\exists x_0 \in A_i$ for in fact every $i \in I$. Now, since each $A_i$ is individually connected, every continuous function $f: A_i \rightarrow \left\lbrace 1, -1 \right\rbrace$ is constant and its value must be $f \left( x_0 \right)$. Now, if we consider any continuous function $f: \bigcup\limits_{i \in I} A_i \rightarrow \left\lbrace 1, -1 \right\rbrace$, it must again be constant because of the argument above. Hence, $\bigcup\limits_{i \in I} A_i$ is connected. $\endgroup$ – Aniruddha Deshmukh Aug 23 '18 at 11:42
  • $\begingroup$ i edited my question can you see it please $\endgroup$ – Vrouvrou Aug 24 '18 at 9:46

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