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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

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  • $\begingroup$ What do you mean by "an arc-connected neighborhood from $x$ to $x$"? $\endgroup$ – tomasz Mar 25 '18 at 19:56
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    $\begingroup$ The first condition is probably supposed to say "the arc components of any open subset are open" $\endgroup$ – tomasz Mar 25 '18 at 19:59
  • $\begingroup$ @tomasz I have already corrected that. $\endgroup$ – user402543 Mar 25 '18 at 20:01
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    $\begingroup$ @user402543 Correct a. $\endgroup$ – William Elliot Mar 25 '18 at 20:07
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(arc-connected is just path-connected in Massey's book ,it's not the slightly stronger version where we need homeomorphs of $[0,1]$).

$(a)\Rightarrow (b)$: Let $x \in X$ and let $O$ be any open subset of $X$ that contains $x$. Then let $C_x$ be the arc-component of $x$ in $O$. By $(a)$ we know that $C_x$ is open, and so we have found an open arc-connected $x \in C_x \subseteq O$, showing that the set of all open arc-connected subsets that contain $x$ is the required local base at $x$, showing $(b)$.

As noted $(b) \Rightarrow (c)$ is trivial, as open neighbourhoods are neighbourhoods in particular.

Likewise $(c) \Rightarrow (d)$ is also trivial. We find an arc-connected $V$ inside $U$ by $(c)$ and then we can connect two points of $V$ even inside $V$, so certainly inside $U$.

$(d) \Rightarrow (a)$: Let $O$ be an open subset of $X$ and suppose that $C$ is an arc-connected component of $O$. Let $x \in C$. We apply $(d)$ to $O$ and $x$ and get a neighbourhood $V$ of $x$ such that any two points of $V$ can be connected by an arc inside $O$. Try to show that $C \cup V$ is arc-connected and this then shows that $V \subseteq C$ as $C$ is maximally arc-connected. So $x$ is an interior point of $C$ and so $C$ is open, showing $(a)$.

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  • $\begingroup$ How do I prove that $C\cup V$ is arc-connected? Taking an arbitrary point in $C$ and another in $V$ and showing that there is a path that connects them? $\endgroup$ – user402543 Mar 26 '18 at 21:49
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Hint: Proceed in the direction $(a)\Rightarrow(b)\Rightarrow(c)\Rightarrow (d)\Rightarrow(a)$. All of these implications are fairly straightforward.

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  • $\begingroup$ What does it mean that "Every point has a basic family of arcwise-connected open neighborhoods".? $\endgroup$ – user402543 Mar 25 '18 at 23:55
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    $\begingroup$ @user402543: it means that at every point, you have a basis of arc-connected open sets. $\endgroup$ – tomasz Mar 26 '18 at 22:38

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