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Give an example of a space which is locally compact at all but one point.

My professor told me that

Wedge product of infinitely many circles

Will do the job. But I'm not getting his answer. Other examples will be appreciated too.

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    $\begingroup$ How about the real line plus the point at infinity with a neighborhood around infinity is a set that contains all numbers with sufficiently high magnitude? Then I don't think the point infinity has a compact neighborhood $\endgroup$ – Mark Mar 26 at 8:10
  • $\begingroup$ Actually, this one came to me also. I'm not sure $\endgroup$ – MathCosmo Mar 26 at 8:11
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    $\begingroup$ @Mark : since $\Bbb R$ is homeomorphic to the open interval $(0,1)$ adding the point $\infty$ with the right half-lines as neighborhoods of it will make the space homeomorphic to $(0,1]$ which is certainly locally compact at every point. $\endgroup$ – Andrea Mori Mar 26 at 8:16
  • $\begingroup$ I was suggesting that each open set of infinity has right and left half lines. But I see that would give homeomorphism to a circle $\endgroup$ – Mark Mar 26 at 8:37
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A variation of the wedge space that your professor suggested: the hedgehog of infinite "spininess":

Let $X = [0,1] \times \mathbb{R}$, where we identify all points $(0,t)$ to a single point (equivalence class, really, we coudl call it plain $0$, the origin) with the metric:

$$d\left((x,t), (x',t')\right) = \begin{cases} |x-x'| & \text{ if } t=t' \\ |x| + |x'| & \text{ if } t \neq t' \\ \end{cases}$$

so the space looks like $\mathbb{R}$ many copies of $[0,1]$ glued together at their $0$, metrically. If does not matter what representation $(0,t)$ we choose for $0$ in this computation (so the metric is well-defined).

At all points $(x,t)$ different from $0$ the space behave locally just like $(0,1]$ so is locally compact, but no neighbourhood of $0$ is compact: for some ball neighbourhood $B(0,r)$ for some $r>0$ pick $0 < s < r$ and note that $\{(s,t): t \in \mathbb{R}\}$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $\mathbb{N}$ in the second coordinate for this...)

For a nice overview at an elementary level of such "hedgehog spaces", see this paper, e.g. They can be quite useful.

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HINT: In slightly more elementary terms what your professor is suggesting is the following.

Consider infinitely many (a numerable infinity will suffice) copies $C_1$, $C_2$, $C_3$, ... of the circle $S^1$ with its natural topology and on each copy fix a point $$ P_i\in C_i. $$ Now consider the quotient $$ X=\left.\bigcup_{i}C_i\right/\sim $$ where the only non trivial equivalence is $P_1\sim P_2\sim P_3\sim\cdots$. Let $P\in X$ be the class of the points $P_i$. What is a neighborhood of $P$?


(Added later)

More generally you can consider a numerable family of pointed locally compact spaces $\{T_i, P_i\in T_i\}_{i\in\Bbb N}$ and form the quotient space $$ X=\left.\bigcup_{i\in\Bbb N}T_i\right/\sim $$ where the $P_i$ are declared equivalent to each other, and $P\in X$ their common class. Then $X$ is locally compact at each of its points $Q\neq P$ but $P$ doesn't have a compact neighborhood since every open set $P\in U\subset X$ is of the form $$ U=\left.\bigcup_{i\in\Bbb N}A_i\right/\sim $$ with $A_i$ open in $T_i$ and $A_i\cap A_j=\{P\}$ for every $i\neq j$.

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  • $\begingroup$ This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks. $\endgroup$ – Bertrand Wittgenstein's Ghost Mar 26 at 8:54
  • $\begingroup$ We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology. $\endgroup$ – Henno Brandsma Mar 26 at 9:02
  • $\begingroup$ @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer. $\endgroup$ – Andrea Mori Mar 26 at 10:13
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    $\begingroup$ @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear $\endgroup$ – Andrea Mori Mar 26 at 10:15
  • $\begingroup$ @AndreaMori Thank you very much. $\endgroup$ – Bertrand Wittgenstein's Ghost Mar 26 at 17:37
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Here is a different example. Take the universal cover of a disk in $\mathbb{R}^2$: $\widetilde{B(0,1)−\{0\}}$. Endow it with the Euclidean metric it inherits from the disk, and then take its completion with respect to that metric.

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