1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ $A\subset B$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.