# Why is there no "weakly version" of locally path-connectedness?

As usual in topology, there are many different definition for terms like "locally connectedness":

Let $$(X,\mathcal{T})$$ be a topological space. Note that I use the following definition of neighbourhood: A set $$U\subset X$$ is a neighbourhood of $$x\in X$$, if there is an open set $$\mathcal{O}\subset X$$, such that $$x\in\mathcal{O}\subset X$$. Therefore, neighbourhoods can also be closed.

(1) $$(X,\mathcal{T})$$ is called "weakly locally connected" at $$x\in X$$, if for every neighbourhood $$U\subset X$$ of $$x$$ there is a connected neighbourhood $$V$$ of $$x$$, such that $$x\in V\subset U$$. In othere words $$x$$ admits a neighbourhood basis of connected sets. If $$(X,\mathcal{T})$$ is weakly locally connected at every $$x\in X$$, then it is called "weakly locally connected".

(2) $$(X,\mathcal{T})$$ is called "locally connected" at $$x\in X$$, if for every open neighbourhood $$U\subset X$$ of $$x$$ there is an open connected neighbourhood $$V$$ of $$x$$, such that $$x\in V\subset U$$. In othere words $$x$$ admits an open neighbourhood basis of connected sets. If $$(X,\mathcal{T})$$ is locally connected at every $$x\in X$$, then it is called "locally connected".

The two definition differ just by the word open.

Obviously, if $$(X,\mathcal{T})$$ is locally connected at $$x\in X$$, it is also weakly locally connected at $$x$$. The reverse is in general not true. However, we can show that every weakly locally connected space is also locally connected. Therefore, the two definition are globally equivalent. (For a proof see https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Locally_Connected_Space)

I am wondering, why there is no such thing as a weakly locally path-connected space. I have found both definition of locally path-connectedness in textbooks: every neighbourhood contains a path-connected neighbourhood and the second one with open neighbourhoods....are they in this case already equivalent locally? Also Wikipedia doesn't make a distinction: They define weakly locally connected and locally connected, but they define only locally path-connected without a weakly version... (https://en.wikipedia.org/wiki/Locally_connected_space)

In other words: Is the following statement true:

Let $$x\in X$$ be fixed.

Every neighbourhood $$U$$ of x has a path-connected neighbourhood $$V$$, such that $$x\in V\subset U$$ $$\Rightarrow$$ Every open neighbourhood $$U$$ of x has an open path-connected neighbourhood $$V$$, such that $$x\in V\subset U$$

That the statement is again globally (if we assume that both sides holds for all x) true is obvious, by a similar proof as for locally conectedness..... But I guess that is also true at a certain point, because otherwise there would also ba such a thing as weakly locally path-connectedness....

• Hi, I would just like to point out that there is no point in writing "open" twice. If every open neighbourhood of $x$ admits an open sub-neighbourhood which path connected (or whatever), then the same is true for every neighbourhood of $x$ (since it is necessarily contains an open neighbourhood anyway). The only important thing is that you're looking for a sub-neighbourhood which is open and path-connected. Jul 5, 2020 at 6:50
• Hi. The wikipedia page has now been edited with the notion of "path connected im kleinen" aka "weakly locally path connected" at a point. The reference to the article by Bjorn et al. in there has a nice diagram in section 2 summarizing the relationship between the properties. Jun 26, 2022 at 23:46
• Thank you very much for letting me know @PatrickR ! :-) Jun 27, 2022 at 13:43

The statement is not true. The space described in this post Consider the "infinite broom" is used to show that weakly locally connected at a point $$x$$ does not imply locally connected at $$x$$, but it works equally well to show that weakly locally path-connected at $$x$$ does not imply locally path-connected at $$x$$.
The space $$X$$ can be described as the subset of $$\mathbb{R}^2$$ consisting of the union $$\bigcup_{n,k\in\mathbb{N}} L_{n,k}$$ along with the line segment joining $$(0,0)$$ and $$(1,0)$$, where each $$L_{n,k}$$ is the line segment joining the points $$(\frac{1}{n+1},\frac{1}{n+1+k})$$ and $$(\frac{1}{n},0)$$. Our distinguished point $$x$$ will be $$(0,0)$$. For any neighborhood $$U$$ of $$x$$, we can find some $$N\in\mathbb{N}$$ such that $$Y_n=\bigcup_{n\geq N,k\in\mathbb{N}} L_{n,k}\cup [0,\frac{1}{N-1}]\times\{0\}$$ is contained in $$U$$. Then $$Y_n$$ is a path-connected neighborhood of $$x$$ (though not open) and thus $$X$$ is weakly locally path connected at $$x$$.
However, we cannot find a path-connected open neighborhood of $$x$$ contained in $$U$$. Suppose $$U$$ is a neighborhood of $$x$$ which does not contain the point $$(1,0)$$ and that $$V$$ is a path-connected open neighborhood of $$x$$. Since $$V$$ is neighborhood of $$x$$, it must contain $$(\frac{1}{n+1},0)$$ for some $$n$$. Since $$V$$ is open, it must also contain $$(\frac{1}{n+1},\frac{1}{n+1+k})$$ for some $$k$$. And since $$V$$ is path-connected and contains $$(\frac{1}{n+1},\frac{1}{n+1+k})$$, it must also contain $$(\frac{1}{n},0)$$. We may continue in this fashion and conclude that $$V$$ contains $$(1,0)$$ and is therefore not a subset of $$U$$. We did this by only assuming that $$V$$ was open, a neighborhood of $$x$$, and path-connected. Thus, $$X$$ is not locally path connected at $$x$$.