Help needed in a proof to show path connected-ness of a special kind of set in real Euclidean space Let $A$ be a connected subset of $\mathbb R^n$ and $\epsilon >0 $ , consider $V:=V(\epsilon , A):=\{x \in \mathbb R^n : d_A(x)< \epsilon  \}$ , I need to show that $V$  is path connected , so we are done if $V$ is open and connected . Now I have shown that $V$ is open . Let $x,y \in V$ , if I can show $x,y \in B $ for some connected $B \subseteq V$ , then we will be done . So let $x,y \in V$ , then $\epsilon$ is greater than the infinimum   of the set $\{d(x,a) : a \in A \}$ , so for some $a \in A , d(x,a) < \epsilon $ and similarly for some $b \in A$ , $d(x,b) < \epsilon$ (here $d$ is the usual euclidean distance on $\mathbb R^n$ ) . Now every point in the sets $[x,a]:=\{cx+(1-c)a : c \in [0,1]\}$ , $[y,b]:=\{cy+(1-c)b : c \in [0,1]\}$ , $A$ lie in $V$ , and all are connected , so I wanted to take 
$B:=A \cup[x,a] \cup [y,b]$ , now I know $A \cap [x,a]$ and $A \cap [y,b]$ are non-empty as they contain 
$a,b$ respectively , but I am unable to show $[x,a] \cap [y,b]$ is non-empty and so I cannot complete my proof to show $B$ is connected . So , how to show $[x,a] \cap [y,b]$ is non-empty ? Please help . Thanks in advance .
 A: To show $V$ is connected use the following facts:


*

*The set in question $V = \displaystyle\bigcup_{a \in A}B(a, \varepsilon)$ where $B(a, \varepsilon)$ is the ball in Euclidean space with centre $a$ and radius $\varepsilon$.

*Each $B(a, \varepsilon)$ is connected.

*If two connected sets (for example $A$ and some $B(a, \varepsilon)$) overlap their union is connected.

*If a family of connected sets has nonempty intersection, the family's union is connected.
Good luck!
A: Your proof is virtually complete as it stands (apologies for writing it out again with slightly different notation in an earlier version of this answer). To show that $X \cup Y \cup Z$ is connected where $X$, $Y$ and $Z$ are connected, you do not need to show that $X \cap Y \cap Z$ is non-empty. All you need is to show that two of $X\cap Y$, $X \cap Z$ and $Y \cap Z$ are non-empty. Think of hanging (connected) shirts on a (connected) washing line: the shirts don't have to touch for the whole configuration to be connected. With $X= A$, $Y = [x, a]$ and $Z = [y, b]$, you have that $X \cap Y$ and $X \cap Z$ are both non-empty, hence your set $B$ is connected and you are done.
