Prove that the following four conditions on a topological space are equivalent:
(a) The arc components of any open subset are open.
(b) Every point has a basic family of arcwise-connected open neighborhoods.
(c) Every point has a basic family of arcwise-connected neighborhoods (they are not assumed to be open).
(d) For every point $x$ and every neighborhood $U$ of $x$, there exists a neighborhood $V$ of $x$ such that $V\subset U$ and any two points of $V$ can be joined by an arc in $U$.
Thus, any one of these conditions could be taken as the definition of local arcwise connectivity.
This exercise is from the book "A basic course in algebraic topology " by Massey, could someone help me please to give meaning to this? How would you prove this?
My definition of being locally arcwise-connected is this: We say that $X$ is arcwise-connected at $x\in X$ if for every open neighborhood $U$ of $x$ there is an arcwise-connected neighborhood $V$ from $x$ to $x\in V\subset U$. If $X$ is locally arcwise-connected for all $x\in X$, we will say that $X$ is locally arcwise-connected, how can I relate this to exercise? Thank you very much. ($ (b) \to (c) $) is trivial