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In the definition of a locally connected space we demand every neighbourhood of a point to satisfy certain condition whereas for a locally compact space we demand that one neighbourhood be there with the required property.

Is there some reason for this difference? Is it so that a compact space needs(?) to be locally compact?

Connectedness is a "geometric" property whereas compactness is an "analytic" property. Is that a reason behind such different definitions?

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It's more general than this -- "locally" is used in both senses depending on what's being talked about, it's not particular to just connectedness and compactness. –  Harry Altman Oct 13 '11 at 5:00
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Not really every neighbourhood, but a local base of them. –  Henno Brandsma Oct 13 '11 at 6:07
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4 Answers

up vote 12 down vote accepted
  1. One could phrase local compactness using neighbourhood bases as well (in the Hausdorff case, at least) if desired: once one has one precompact open neighbourhood, one automatically has a whole neighbourhood base of precompact sets, since any subset of a precompact set is still precompact. (And in practice, this is often how local compactness is actually used.)

  2. The most important thing that a property named "locally P" should obey is that it be local rather than global. In the case of topological spaces, this means that if a topological space X is locally P at some point $x_0$, and we have another topological space Y which agrees with X at a neighbourhood of $x_0$ (e.g. Y could be the restriction of X to a neighbourhood of $x_0$, with the relative topology), then Y should be locally P at $x_0$ as well. Both local compactness and local connectedness, as defined traditionally, have this locality property. On the other hand, the property "$x_0$ has at least one connected neighbourhood" is not local (consider for instance $({\bf Q} \times {\bf R}) \cup \{\infty\}$ in the Riemann sphere ${\bf R}^2 \cup \{\infty\}$, which is a globally connected space which is locally identical near the origin to ${\bf Q} \times {\bf R}$ in ${\bf R}^2$, which has no connected open sets), and thus this property does not deserve the name of "local connectedness at $x_0$".

It may help to think of the modifier "locally" not as a rigid recipe for converting global properties to local ones, but rather as an indicator that the property being modified is a local analogue of the global, unmodified, property. In most cases there is only one obvious such analogue to select, although in some cases (such as the notion of "locally compact" in the non-Hausdorff setting) there is some freedom of choice, which ultimately means that one has make a somewhat arbitrary convention regarding terminology at some point.

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This is a very neat way to justify my intuitive but not so elaborate and well justified belief that any neighborhood should have the property. Note taken! :-) I think of a neighborhood as a "sufficiently large set". Large enough to be considered a neighborhood. Often, one needs to choose a "small" set, but at the same time, big enough to make the small step big enough to achieve the goal... "path connect two points", for instance. –  André Caldas Oct 13 '11 at 18:57
    
Thank you Terry for the explanation. Now I guess I understand the modifier "locally" better. –  Shripad Oct 14 '11 at 7:24
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If $X$ is a space and "P" is an adjective that can apply to spaces, then in many cases "$X$ is locally P" means "for every point $x\in X$, for every neighborhood $N$ of $x$, there is a neighborhood $N'$ of $x$ in $X$ such that $N'\subset N$ and $N$ is P", or more briefly "every point of $X$ has arbitrarily small P neighborhoods". I would agree with Terry, though, in cautioning against a strict rule.

This gives the usual notion of locally connected (which turns out to be equivalent to saying that every component of every open set is open), and likewise for locally path-connected. It also gives one of the usual notions of locally compact, generally considered better than the other usual notion, which happens to agree with it in the Hausdorff case so that the difference is often irrelevant.

Here is one way in which people confuse each other with "locally" and "local". What do I mean if I say that a map $f:X\to Y$ is locally a homeomorphism? Surely I do not mean that every $x$ has arbitrarily small neighborhoods $N$ such the the restriction to $N$ is a homeomorphism $N\to Y$. What I mean is that every $x$ has arbitrarily small neighborhoods $N$ such the the restriction to $N$ gives a homeomorphism $N\to f(N)$. OK, so this conforms to the general rule of my first paragraph [extended to the case where "P" is allowed to be "a homeomorphism" even though that's not an adjective] if I give "homeomorphism" the old-fashioned meaning of "homeomorphism to its image". (Also I could have omitted "arbitrarily small", but that's beside the point.)

But that's not the confusion that I was thinking of. Here it comes. I have known more than one student to misinterpret the expression "locally homeomorphic" in the following way: you learn what a local homeomorphism is (perhaps while studying covering spaces), and then later someone says something about $X$ being locally homeomorphic to $Y$. And you assume that this means that there is a map $f:X\to Y$ that is a local homeomorphism, whereas you should have been thinking "every $x$ has arbitrarily small neighborhoods $N$ such that $N$ is homeomorphic to $Y$".*

*or more likely such that $N$ is homeomorphic to a neighborhood of a point in $Y$. See, that's an example of what Terry is saying.

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I, myself, am revolted with such a definition, too.

It seems, according to Wikipedia, that there is no consensus on the definition. And probably, few people care about it, because in Hausdorff spaces, all the definitions are equivalent. In addition to the definition you present, one could also say that a space is locally compact when every point has a closed compact neighbohood. In general, even if a neighborhood is compact, it does not mean its closure will be compact as well.

In the Hausdorff case, the closure of subsets of compact sets are compact, since every compact set is closed in this case. Also, in the Hausdorff case it is true that if $K$ is a compact neighborhood of $x$, then $x$ has a neighborhood filter base made out of compact sets. This is because $K$ is a normal space.

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Wow, I never knew this! I always figured the strange definition was to give nice properties, but in fact the nice properties are all for locally compact Hausdorff spaces (e.g. they are characterized by being open subspaces of compact Hausdorff spaces), so I'm guessing point-set topology courses everywhere could just use the standard "locally <blank>" definition and avoid this confusion completely. –  David White Oct 13 '11 at 13:58
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Which one is your "standard 'locally <blank>' definition"? :-) –  André Caldas Oct 13 '11 at 14:52
    
For all points p and for all neighborhoods U there is a smaller neighborhood V satisfying the property. I think this is the standard for every locally <blank> other than locally compact –  David White Oct 13 '11 at 22:05
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I think the best definition of locally compact is that the lattice of open sets is a continuous lattice in the sense of Dana Scott. This is what is needed for a space X to have the property X\times() is adjoint to Hom(X, ). See Johnstone's Stone Space book for more on this.

Added I will flesh out this answer now that I have more time. In a lattice, an element $x$ is said to be way below an element $y$, written $x\ll y$, if whenever $y$ is below a join $\bigvee A$, then $x$ is below $\bigvee F$ where $F$ is a finite subset of $A$. (This kind of means $x$ is compact in $y$). If $T$ is a topological space and $\mathcal O(T)$ is the lattice of open sets, then $U\ll V$ iff there is a compact space $K$ with $U\subseteq K\subseteq V$.

A complete lattice is a continuous lattice if each element is a directed join of elements way below it. It is not hard to prove that $\mathcal O(T)$ is continuous for a Hausdorff space iff $T$ is locally compact. It has been argued that this is the right general definition of a locally compact space. The main positive evidence, if memory serves, is that $\mathcal O(T)$ is continuous iff for all spaces (maybe some minor separation condition like sober is needed??) $Y,Z$ one has $Hom(T\times Y,Z)\cong Hom(Y,Hom(T,Z))$ (i.e. $T$ is exponentiable). I think this result holds on the nose in the context of locales (pointless spaces).

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Is $X=T$? –  Emil Jeřábek Oct 14 '11 at 10:29
    
Yes, I'll fix it. –  Benjamin Steinberg Oct 14 '11 at 10:47
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