How can we formally show that this holds true:
A locally path connected space is connected iff if it is path connected? True or False?
I think it is true, however I seem to be confused on how this can be shown. I would appreciate some nice way to show how the above statement is true.
Assume $S$ is connected and locally path connected. For $a,\ b$ in $S$, there is a finite overlaping chain of base sets $\{ U_1, U_2,\cdots, U_n \}$ with $a \in U_1,\ b \in U_n$ and consecutive sets intersecting (since $S$ is connected).
Find some $x_j \in U_j \cap U_{j+1}$ for $j = 1\cdots n-1$. Since the $U_j$'s are path connected, construct a path from $ a$ to $x_1$ to $x_2$ to $\cdots$ to $x_{n-1}$ to $b$.
Recall that any topological space can be partitioned into its path components and the space is path connected when it consists of exactly one path component.
Let $X$ be a locally path connected. Its path components are easily seen to be open. We can therefore argue
If $X$ is not path connected let $U$ be be the open path component containing some points in $X$ with a nonempty complement. But the complement $V$ of $U$ is the union of open path components and must be open. But then $X$ has a nontrivial decomposition
$\quad X = U \cup V$
into two open sets and therefore $X$ is not connected.
