Comparing definitions local connectedness and local compactness

Question

So after comparing the definitions, I have no idea why these seem to be defined so differently.

A space $X$ at $x$ is locally connected if there is an open set $U$ containing $x$ such that there we can find a open connected set $x \in V \subset U$. I guess in other words, the open connected sets $V_x$ form a local base.

A space $X$ at $x$ is locally compact if there is a compact neighbourhood $C$ at $x$.

It looks like to me that local compactness is roughly saying a space $X$ is locally compact on some point $x$, if on some neighbourhood $U(x)$, $X$ is compact on that neighbourhood. In other words, $U(x) \cap X = U(x)$ is compact, which seems to agree with what one might normally come up with for the definition.

On the other hand, local connectedness says we can find a connected neighbourhood inside an open set...

I've read some answers from MO, but none of these feels satisfying to me (a lot of red-herring answers I feel...). I just don't understand what's wrong with defining local connectedness as having "a connected neighbourhood" or local compactness as "a compact set inside an open set". Maybe there are spaces that simply do not have connected (or compact) subsets?

Answer 1

More precisely, locally compact means has a base of open sets with compact closures. A weaker definition is: has a base of open sets each of which is contained in a compact set. For Hausdorff spaces, the definitions are equivalent. Thus is a definition in the form of has a base of open sets that are whatever; like locally connected - has a base of open connected sets.

Answer 2

Normally you would define local compactness like this:

For every point $x$ and a neighbourhood $x\in U$ there is a subneighbourhood $x\in V\subset U$ such that $\overline{V}$ is compact.

The thing is that these two definitions of local compactness are equivalent. Indeed, if $U$ is relatively compact (i.e. $\overline{U}$ is compact) and $V$ is any other subset then $U\cap V$ is again relatively compact.

The analogy does not work for connectedness. The intersection of a connected open subset with any other open subset does not have to be connected (even when both are connected).

Let's define:

• C: connected
• LC: locally connected (with the usual definition)
• LC2: every point has a connected neighbourhood

Then C implies LC2. Indeed, if $X$ is connected then every point has a connected neighbourhood, namely $X$. Also LC implies LC2.

On the other hand neither LC implies C (discrete space with at least two points) nor C implies LC (the topologist's sine curve).

As for why LC is used instead of LC2, well it's because "necessity is the mother of invention".

Answer 3

In general, if $P$ is a property of topological spaces, we say that $X$ satisfies $P$ locally at $x\in X$ if $x$ has a basis of neighbourhoods satisfying $P$, just as you say when discussing local connectedness, and we say that $X$ satisfies $P$ weakly locally at $x$ if $x$ has a neighbourhood satisfying $P$.

Then, $X$ is (weakly) locally $P$ if it is such at every $x\in X$.

Now, in general, the property $P$ trivially implies the weakly local property $P$ (since we can take all of $X$ as the neighbourhood witnessing weakly local $P$), but not the local property $P$.

For example, if $X_0$ is a totally disconnected space without isolated points (like the rationals, $\mathbf Q$), then the cone of $X_0$, i.e. $X=X_0\times [0,1]/{\sim}$, where $(x,t)\sim (x',t')$ when $(x,t)=(x',t')$ or $t=t'=1$ is a space which is connected (even path-connected), and thus weakly locally connected, but not locally connected, since for points with $t<1$, neighbourhoods look like pieces of $X_0\times[0,1]$, which are certainly not connected.

Similarly, if $X_0$ is a space such that no point has a compact neighbourhood (again, like $\mathbf Q$), we can embed $X_0$ in a silly compactification $X=X_0\cup \{\infty\}$, topologized so that all open substets of $X_0$ are open in $X$, and the only neighbourhood of $\infty$ is $X$ itself.

Then $X$ is likewise weakly locally compact (because it is compact), but not locally compact.

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However, if $X$ is a Hausdorff space, then it is weakly locally compact if and only if it is locally compact.

To see this, consider any $x\in X$ and let $U$ be a compact neighbourhood of $x$. Then, since $X$ is Hausdorff, $U$ is necessarily closed in $X$ (because a compact subset of a Hausdorff space is closed) and compact Hausdorff.

It follows that $U$ is completely regular and so $\{x\}$ is the intersection of closures of the open neighbourhoods of $x$ in $U$, and hence in $X$ (because $U$ is closed in $X$). But these closures are compact as closed subsets of the compact space $U$, so they form a basis of compact neighbourhoods of $x$ in $X$.

Because of this, when one considers only Hausdorff spaces, weakly local compactness is used in place of local compactness, because it is simpler to verify and understand.