# On the definition of locally connectedness.

Why don’t we just define the locally connectedness the same way we define locally compactness, that is, every point has a connected neighborhood?

On the wiki page the weak locally connectedness and connectedness are proved to be “almost identical”, but it does not mention the motivation of the definition.

## 3 Answers

Local compactness is an easier case: we mostly consider that property in the context of Hausdorff spaces, and for Hausdorff spaces it's equivalent to demand one compact neighbourhood or a local base of compact neighbourhoods in the definition.

The latter defintition of localisation of a property is more intuitive: however "close" to a point we are, we "see" compactness. In proofs we often need a compact neighbourhood inside some open set we are working in, e.g. and the local base variant gives us lots of options. One defines certain notions to prove results, and we get better results with a local base variant.

One one the reasons to study local (path-)connectedness is to make a distinction between different kinds of connected spaces. The "one connected neighbourhood" definition would kill that idea as then a connected space is locally connected automatically, chosing $$X$$ as the connected neighbourhood. There even are more refined notions of local connectedness like "connected im Kleinen" (somewhat old-fashioned nowadays) to distinguish among locally connected spaces in the strong sense.

I think the crucial fact is that (reference)

A space is locally connected if and only if for every open set U, the connected components of U (in the subspace topology) are open.

With the other definition (the weak one), this would be false.

Consider a space like the closed topologist's sine curve, i.e., the closure $$\mbox{cl}(G)$$ where $$G$$ is the graph of $$\sin(1/x)$$ for $$0 < x \le \pi$$. This space is connected, so every point has a connected neighbourhood, namely the whole space, but has points (in this case the points that lie on the $$y$$-axis) for which all sufficiently small neighbourhoods are disconnected.

• I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $x\in\mathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood. – William Sun Dec 4 '18 at 21:39
• Sorry. I said that wrong. I've fixed the answer. – Rob Arthan Dec 4 '18 at 21:40