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 $\mathbb{C}$?
Please need help.......
 A: 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.
A: 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. 
