Vtyjkyuk

Please is this prof is correct ?
Let $Omega_i_iin I$ a famille of connected sets such that $$forall i,jin I, Omega_icapOmega_jneqemptyset$$
I want to prove that $bigcup_iin I Omega_i$ 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
$$
begincases
bigcup Omega_i= Acup B\
Acap B=emptyset
endcases
$$
we have $forall iin I, Omega_isubset bigcup_iin IOmega_i=Acup B$ then by the connectedness of $Omega_i$
$$forall iin I, [Omega_isubset A ~textor~ Omega_isubset B]$$

As $forall i,jin I, A_icap A_jneq emptyset$ we deduce that $$forall Iin A, Omega_isubset A~textor~ forall iin I,Omega_isubset B$$
it follows that $B=emptyset$ or $D=emptyset$, which is a contradiction.

Please if I change the condition
$$forall i,jin I, Omega_icapOmega_jneqemptyset$$
by
$$
exists i_0in I, Omega_i_0cap Omega_jneq emptyset,forall jin I
$$
How to do ?
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $Omega_i_iin I$ a famille of connected sets such that $$exists i_0in I, Omega_i_0cap Omega_jneq emptyset,forall jin I$$
I want to prove that $bigcup_iin I Omega_i$ 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
$$
begincases
bigcup Omega_i= Acup B\
Acap B=emptyset
endcases
$$
we have $forall iin I, Omega_isubset bigcup_iin IOmega_i=Acup B$ then $Omega_i_0subset Acup B$ as it is connected we have $$Omega_i_0subset A~textor ~ Omega_i_0subset B$$ if we suppose that $Omega_i_0subset A$ then $$forall jin I, Omega_jcap Aneq emptyset $$ by the connectedness of $Omega_i$ we deduce that $$forall jin I, Omega_jcap B=emptyset$$ then $$forall jin I, Omega_jsubset A$$ thus $$bigcup_jin IOmega_jsubset A$$ so $B=emptyset$ contradiction in the same way if we suppose that $Omega_i_0subset B$ we find that $A=emptyset$

thank you

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 $f: X rightarrow leftlbrace 1, -1 rightrbrace$ is constant.

Your second case can be easily proven by this fact. Notice that each $Omega_i$ is connected and hence every continuous function $f_i: Omega_i rightarrow leftlbrace 1, -1 rightrbrace$ 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 $f : bigcuplimits_i in I Omega_i rightarrow leftlbrace 1, -1 rightrbrace$ is constant and hence $bigcuplimits_i in I Omega_i$ is connected.