Definition of locally pathwise connected

Question

I have two definition of this property.

Def 1: A space X is said to be locally path connected at x if for every (open) neighborhood U of x, there is a path-connected (open) neighborhood V of x contained in U. If X is locally path connected at each of its points, then it is said to be locally path connected.

Munkres, Topology

Def 2: A space X is locally pathwise connected if each point has a nhood base consisting of pathwise connected sets.

Willard, General Topology

Notice, Def 1 is speaks of an open neighborhood, Def 2 of a generic neighborhood.

Are the two definition equivalent, if not which one is preferable?

Answer 1

Munkres regards all neighborhoods as open (see the definition on p.91). I do not know whether Willard has the same understanding of neighborhoods, but if so, then both definitions are obviously equivalent because a neigborhood base of $x \in X$ is a collection $\mathcal{U}$ of neigborhoods of $x$ such that each neigborhood $V$ of $x$ contains some $U \in \mathcal{U}$.

Added: I just read the proof of Theorem 25.3 in Munkres which can be "copied" to obtain a proof of Theorem 25.4.

The proof shows that the same results are true if we understand a neighborhood of $x$ to be any set $N$ such that $x \in \text{int}(N)$. In fact, look at the first part of proof. Then we get $x \in \text{int}(V) \subset V \subset C$ which suffices to see that $C$ is open.

Answer 2

As shown in my comment, the two definitions are equivalent for local pathwise-connectedness of the whole space. The corresponding definitions for local pathwise connectedness at a point are not equivalent. The example given here Consider the "infinite broom" works for local pathwise connectedness as well as local connectedness. In that context, the definition including "open" is preferable.

Answer 3

Here is a more general result.

Let us understand a neighborhood of $x$ to be any set $V$ such that $x \in \text{int}(V)$.

A space $X$ is said be locally $0$-connected, written as $LC^0$, if for each $x \in X$ and each neighborhood $U$ of $x$ there exists a neigborhood $V$ of $x$ such that $V \subset U$ and such that for each $y \in V$ there exist a path in $U$ connecting $x$ and $y$.

See for example Chapter IV 8.13 in

Dold, Albrecht. Lectures on algebraic topology. Springer Science & Business Media, 2012.

$LC^0$ seems to be weaker than locally path connected because in the above definition the path connecting $x$ and $y$ is not required to stay in $V$. However

Theorem. The following are equivalent:

(1) All path components of open sets are open.

(2) Each $x \in X$ has a neighborhood base consisting of pathwise connected open sets. [Munkres]

(3) Each $x \in X$ has a neighborhood base consisting of pathwise connected sets. [Willard]

(4) $X$ is $LC^0$.

(5) For each $x \in X$ and each neighborhood $U$ of $x$, the path component of $x$ in $U$ is a neighborhood of $x$.

Proof. (1) $\Rightarrow$ (2) : Let $U$ be an open neighborhood of $x \in X$ and let $C$ be the path component of $U$ containing $x$. But $U$ is open, hence (2) is satisfied.

(2) $\Rightarrow$ (3) $\Rightarrow$ (4) : This is obviuos.

(4) $\Rightarrow$ (5) : Let $U$ be a neighborhood of $x \in X$ and $C$ be the path component of $U$ which contains $x$. Choose a neighborhood $V$ of $x$ such that $V \subset U$ and such that for each $y \in V$ there exists a path in $U$ connecting $x$ and $y$. Then clearly $y \in C$. Hence $V \subset C$.

(5) $\Rightarrow$ (1) : This is obvious.

Note that we could define variants of $LC^0$ by requiring one or both of $U,V$ to be open. The above proof is valid for all these variants, therefore they are equivalent.