We'll consider The Hedgehog space, and use the definition from Wikipedia (link to the current revision):
Let $\kappa$ a cardinal number, the $\kappa$-hedgehog space is formed by taking the disjoint union of $\kappa$ real unit intervals identified at the origin. Each unit interval is referred to as one of the hedgehog's spines. The hedgehog space is a metric space, when endowed with the hedgehog metric $d(x,y)=|x-y|$ if $x$ and $y$ lie in the same spine, and by $d(x,y)=x+y$ if $x$ and $y$ lie in different spines. Although their disjoint union makes the origins of the intervals distinct, the metric identifies them by assigning them 0 distance.
Now, I want to prove that this space is locally compact. For each $t\in [0,1]$, I will denote $t_\alpha$ the belonging to $\alpha$-th spine, where $\alpha \in\kappa$ ($\alpha$ cardinal).
Take an arbitrary $t_{\alpha}$. If $t_{\alpha}\neq 0_{\alpha}$, then $0_{\alpha}<t_{\alpha}\le 1_{\alpha}$. Consider $s\in [0,1]$ such that $0<s<t$. Can we say that $[s_{\alpha},1_{\alpha}]$ is a compact neighborhood of $t_{\alpha}$?
And what about if $t_{\alpha}= 0_{\alpha}$?
Thanks!
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$\begingroup$I don't think this space is locally compact. Indeed the whole space isn't compact (when $\kappa$ is infinite), as the open cover $$\big(B((2/3)_{\alpha},1/2)\big)_{\alpha\in\kappa}\cup\big(B(0,1/2)\big)$$ clearly admits no finite subcover. If the space were locally compact, $0$ would admit a compact neighborhood $K$ with $\overline{B(0,r)}\subset K$ for some $r>0$. However, $\overline{B(0,r)}$ is clearly homeomorphic to the total space which isn't compact, which is a contradiction.
$\endgroup$ $\begingroup$I don't think your space is locally compact if $\kappa$ is infinite. For example, take $\kappa=|\mathbb N|$, meaning you have intervals $I_1,I_2,\dots$ ($I_i$ marks the unit square), joined at $0$. Now take the point $0$ and any neighborhood of $0$. This neighborhood includes a ball $B(0,\epsilon)$ for some epsilon. Now take the covering of $B(0,\epsilon)$ such that $U_0 = B(0,\epsilon/2)$ and $U_i = (\epsilon/4, \epsilon)$ where $U_i\subseteq I_i$. There is no finite subcovering.
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