Apparently the definition of local compactness is not standard. Compactness is obviously a global property of a topological space, so it makes sense to define it locally. Second-countable spaces have countable bases (a global phenomenon), and first-countable spaces are the local version of that, where each point has a countable local base. Being connected is a global property, and being locally connected at a point means every neighborhood is a supserset of some connected set containing the point.
So surely there's no ambiguity as to what a locally compact space should be, right?
So we're all on the same page, a neighborhood is not necessarily open. A relatively compact set is a set whose closure is compact. Potentially the simplest definition of a locally compact space is...
- Every point has some compact neighborhood
And it's obvious that a compact space is locally compact, as itself serves as a compact neighborhood for every point. However, it wouldn't be an analogous notion to locally connected or the first vs second countable spaces, as we're only asserting the existence of one neighborhood, which wouldn't form a sort of local base.
Given this compact neighborhood $K$, we can take any potentially smaller neighborhood $V$ which won't be compact. You may think that because $V \subseteq K$, we can say $\cl{V} \subseteq \cl{K} = K$, and so $\cl{V}$ is a closed neighborhood, subset of a compact space, so it's compact and we're done! But not so fast, $K$ is compact, but not necessarily closed. $\cl{K}$ could be a much bigger set, and so $\cl{V}$ is not necessarily a subset of $K$. We don't truly have a local base. We can certainly fix this!
- Every point has a closed compact neighborhood.
Boom! Now, for any neighborhood $V$ inside this big closed compact one, $\cl{V}$ will be a closed neighborhood that's a subset, and hence also compact. This will indeed form a local base in general, and in this case, each $V$ is "relatively compact" by our definition above. Thus, an equivalent definition of version 2 is "every point has a local base of relatively compact neighborhoods".
Again, the closure of these neighborhoods are compact, but that doesn't mean their closures form a local base, so we can't say that we have a local base of compact neighborhoods.
- Every point has a local base of compact neighborhoods.
Conversely, because compact neighborhoods aren't necessarily closed, we can't argue that their closures are compact. Yes, surprisingly, a compact set may not be relatively compact! Thus, the notions 2 and 3 are not comparable. Much like the leap from 1 to 2, we can leap from 3 to the following.
- Every point has a local base of closed compact neighborhoods.
Now indeed, because they're closed and compact, they're also relatively closed and 4 obviously implies 3. 4 also obviously implies 2 if we just take one of the neighborhoods. Everything implies 1, which makes it the weakest condition.
The problem here is that if a space is compact, it follows that 1 and 2 are true, but 3 and 4 are not necessarily true. Because of this, my take is: 1 should be called weakly locally compact and 2 should be the true locally compact (this way, compact implies locally compact, as it should be). I'm not certain on what to call 3. I'd love to call it strongly locally compact, but this name would sound like 3 implies 2. Maybe truly locally compact, that sounds cool. And 4 is actually known in literature as "locally compact regular" (because 4 actually implies the space is regular), so there you go.