I came across this definition

Let $U\subset S\subset C$. We say that $U$ is relatively open in $S$ if for every $z_0 \in U$, there is $r > 0$ such that $$D(z_0 ;r)\cap S\subset U$$:

Now the author doesnt specify if the subset symbol means that you do not count the trivial subsets ( the empty set and the set itself).

So then i came across another definition.

Definition :A set $S \subset C$ is called connected if the only relatively open and closed sets in S are the empty set and S.

Then i thought take the union of 3 disconnected open balls in C call it $K$.I know this set is disconnected so i want to prove it by the definition above.All i have to do is a find a relatively open closed subset in it except the trivials.SO there must be a closed subset call it $L$ of this set that is relatively open in it Because it is not connected.This means that for every $z_0$ in this closed set there exist an $r>0$ such that if i take the r-ball $\cap K\subset L$ .

Now i am trying to find a pictorial way for it.I drew 3 open circles.And i try to find a closed subset that for any element in it the $r$-ball is inside the closed set.there is no problem for the elements inside the closed set.But at the boundary of my closed subset cant find an $r$-ball without containing some elements that belong to $K$ and not in my closed set so the ball cant be a subset of my of my closed set.

  • $\begingroup$ I do not understand what your question exactly is. If you have a topology on a set $X$ and you have a subset $Y \subseteq X$, then you can define the relative topology on $Y$ as: the open sets of $Y$ are exactly the intersections with $Y$ of the open sets of $X$. These are called "relatively open sets", because otherwise you might confuse them with the open sets of $X$. The same holds for closed subsets. Why should such an $L$ exist? $\endgroup$ – 57Jimmy Jul 27 '16 at 16:46
  • $\begingroup$ You have defined a decomposition of $K$ into a union of 3 pairwise disjoint open balls $$K = B_1 \cup B_2 \cup B_3$$ You seem worried about the points on the boundaries of the balls $B_1$, $B_2$, and $B_3$. But since those points are not elements of the set $K$, there is nothing to worry about. $\endgroup$ – Lee Mosher Jul 27 '16 at 17:10
  • $\begingroup$ I want to prove that the set of 3 disjoint open balls is not connected using the definition of connectedness i said. And try imagine a pictorial way.SInce the et of 3 disjoint open balls is disconnected there must be a closed relatively open set inside the 3 disjoint open balls excepte the trivial subsets. $\endgroup$ – Manolis Lyviakis Jul 28 '16 at 10:01
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    $\begingroup$ $A\subset B$ always means that there is no member of $A$ that does not belong to $B$. It does not require that $A$ is not empty and does not require that $A\ne B.$ The symbols $\subset $ $ \; \subseteq \; $ $\subseteqq \; $ all mean the same thing. To say that $A$ is a subset of $B$ and $A\ne B , $ you write $ A\subsetneq B $ or $ A\subsetneqq B . $ (A\subsetneq B or A\subsetneqq B.) $\endgroup$ – DanielWainfleet Jul 30 '16 at 15:12

Note that each one of the three balls is relatively open AND CLOSED in $K$, being the intersection of a closed ball of $\mathbb{C}$ with $K$. For instance, $B_1 = \overline{B_1} \cap K$, where the closure is meant in $\mathbb{C}$. So you have found a subset of $K$ (namely, $B_1$) which is relatively closed and open in $K$, hence $K$ is disconnected.

Your definition of relatively closed sets is equivalent (in $\mathbb{C}$ and in any other metric space, if with $D(z_0;r)$ you mean the open ball of radius $r$ around $z_0$) to the one I gave in a previous comment (you could try to show this), which is probably easier to visualize.

  • $\begingroup$ I did not understand why my balls are closed.Since the set $K$ is the join of 3 OPEN disjoint balls. $\endgroup$ – Manolis Lyviakis Jul 28 '16 at 10:47
  • $\begingroup$ Your balls are relatively closed, which means closed in the relative topology on $K$. In fact, $B_1$ is the complement, in $K$, of $B_2 \cup B_3$, which is open, hence $B_1$ is certainly closed in $K$, which does NOT mean that it is closed in $\mathbb{C}$! You must pay attention and distinguish between the topology on $\mathbb{C}$ and the relative topology on $K$. $\endgroup$ – 57Jimmy Jul 28 '16 at 10:58
  • $\begingroup$ relative closed set doesnt mean it is closed. The definition says if there is a closed relative open set in K. not a relatively closed which is relatively open. I could argue that the complement of $B_2$ is closed because $B_2$ is open Since you proved $B_1$ is closed and $B_3$ is not a subset of $B_1$ so $B_3$ is closed . so $B_2 \cup B_3$ is not open. $\endgroup$ – Manolis Lyviakis Jul 28 '16 at 11:08
  • $\begingroup$ I think you misunderstood the definition of connectedness: " relatively open and closed sets" means "sets that are both relatively open and relatively closed" (trust me, it is so, it is the standard definition). $\endgroup$ – 57Jimmy Jul 28 '16 at 11:18
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    $\begingroup$ I'll sum it all up for you: a topological space $X$ is connected iff it cannot be written as a disjoint union of two proper closed subspaces iff it cannot be written as a disjoint union of two proper open subspaces iff it does not contain any proper nonempty subspace that is both closed and open (these are all equivalent conditions, and proper means $\subset$ instead of $\subseteq$). This is the general definition. Your definition just says that a subspace of a topological space is called connected iff it is connected in its relative topology, which I have defined in a previous comment. $\endgroup$ – 57Jimmy Jul 28 '16 at 11:26

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