# Connectedness in $\mathbb{C}$

$S=\{i, \frac{i}{2}, \frac{i}{3},...\}$ then $S$ is connected?

"My textbook says, if we join any two points of $S$ by polygonal path , then there are points on this path which do not belongs to $S$ . Thus $S$ is not connected".

"Yes I can see too, there are points on path( between any two points on $S$) which do not belongs to $S$. But is that mean not connected? According to me, this only means, not polygonally connected! But how they concluded not connected?

Is, not polygonally connected⇒not connected ?

If not then, is my textbook solution is wrong?

In perticular, how to check connectedness in $\mathbb{C}$?

• What does "connected" mean? Or, better yet, what does "not connected" mean? Mar 28 '18 at 13:34
• @umberto.p sir, According to my textbook of complex analysis, "an open set $S$ is said to be connected if any two points of set can be joined by a path consisting of straight line segment". Though I also know general definition of connectedness in general topological space. Is polygonally connected is something special? What implications we have if set is polygonally connected? and what implications we have if set is not polygonally connected? Mar 28 '18 at 13:43
• According to the definition you provided, your book is using "connected" to mean what you are calling "polygonally connected". Its proof that $S$ is not connected is what you deem a proof that it is not polygonally connected. It seems your confusion is just an issue of terminology. Mar 28 '18 at 13:58
• Sir, but I need to know what implications follows from polygonally connected? and what implications follows is set is not polygonally connected? Mar 28 '18 at 14:00

Your objection makes sense. Although $S$ is indeed not connected, just asserting that it is not polygonally connected isn't enough to prove it.

You can prove that it is not connected observing that the set $\{i\}$ is a subset of $S$ which is both open and closed. It is closed because it is already a closed subset of $\mathbb C$ and it is open because $B\left(i,\frac12\right)\cap S=\{i\}$. Since $\{i\}\neq\emptyset$ and $\{i\}\neq S$, this proves that $S$ is not connected.

• Thanks sir. then in $\mathbb{C}$, how to connectedness? Or how to show set is not connected? Mar 28 '18 at 13:38
• Sir, further what implications we have from polygonally connected? Mar 28 '18 at 13:45
• @AkashPatalwanshi I've added another paragraph to my answer. Is it all clear now? Mar 28 '18 at 15:38
• Sir, You make it very simple, I get, what you want to say, $S= \{i\}∪\{i\}^c$ hence by definition, it is disconnected. Thank you...:-) Mar 28 '18 at 16:27

There are three notions of connectedness here:

• Polygonally connected: any two points are joined by a path of straight line segments. This notion only makes sense in a vector space.

• Path connected: any two points are joined by a continuous curve.

• (Topologically) connected: A set $S\subset X$ is connected if there do not exist two open sets, $U$ and $V$ of $X$, such that $U\cap V\cap S=\varnothing$ and $S\subset U\cup V$. (If $U$ and $V$ exist, $S$ is said to be separated by $U$ and $V$).

These are inequivalent notions. We have

polygonally connected $\Rightarrow$ path connected $\Rightarrow$ connected,

but in general,

connected $\not\Rightarrow$ path connected $\not\Rightarrow$ connected

For example, the unit circle in $\mathbb C$ is path connected, but not polygonally connected. Examples of connected sets which are not path connected are hard to think of, see the topologist's sine curve.

Your example is disconnected under all three of these notions. To prove this, it suffices to find two separating sets. Let $U_k$ be the open ball of $\frac12(1/k-1/(k+1))$ around $i/k$. The sets $U_k$ are pairwise disjoint and open. Letting $V=U_2\cup U_3\cup\dots$, then $U_1$ and $V$ separate $S$.

One final note: if you restrict your attention to open subsets of $\mathbb R^k$, then all three notions of connectedness are equivalent.

• Thank you so much sir... Mar 28 '18 at 14:26
• Sir, in definition of topological connected, isn't there must be, $U∩V∩S= ∅$ and $S=(U∩S)∪(V∩S)$? Mar 30 '18 at 4:01
• @AkashPatalwanshi The conditions you listed are equivalent to mine. Mar 30 '18 at 12:44
• Again thank you sir, for making it clear...:-) Mar 30 '18 at 12:46
• @AkashPatalwanshi No problem, have a good day :) Mar 30 '18 at 12:47