---- > [!definition] Definition. ([[locally compact]]) > A [[topological space]] $X$ is said to be **locally compact at $x$** if there exists a [[compact]] [[subspace topology|subspace]] $C$ of $X$ that contains an [[neighborhood|open neighborhood]] $U \ni x$. If $X$ is locally compact at all points, we say $X$ is **locally compact**. > \ > Note that any [[compact|compact space]] is automatically locally compact. > [!note] Remark. > At this level of generality the definition is not particularly useful, and in particular does not seem to parallel the 'local' versions of other definitions. But for [[Hausdorff space|Hausdorff]] spaces we get a nice characterization: [[local compactness characterization for Hausdorff spaces]]. ^note > [!basicexample] > - $\mathbb{R}^{\text{std}}$ is [[locally compact]], for given $x \in X$ and $\varepsilon>0$ the compact subspace $[x-\varepsilon, x+\varepsilon]$ containing $x$ contains the open [[neighborhood]] $(x-\varepsilon, x+\varepsilon)$ of $x$. More generally, every [[order topology|simply ordered set]] having the [[least upper bound property]] is [[locally compact]], every point is contained in a [[basis for a topology|basis element]] is contained in a ([[compact]]) [[closed interval]]. > > >- $\mathbb{R}^{n}$ is also locally compact, for analogous reason: [[closed set|closed balls]] about $p \in \mathbb{R}^{n}$ are compact and contain open balls about $p$. > [!basicnonexample] > The [[sequence]] space $\mathbb{R}^{\mathbb{N}}$ is not [[locally compact]]; *none* of its [[basis for a topology|basis elements]] (and hence no open sets, [[open sets are unions of basis elements|which are unions thereof]]) are contained in [[compact]] [[subspace topology|subspaces]]. For if $B=(a_{1},b_{1})\times \dots \times (a_{n}, b_{n}) \times \mathbb{R} \times \dots \times \mathbb{R} \times \dots$ were contained in a compact subspace, [[closure and product|then its]] [[closure]] $\overline{B}=[a_{1},b_{1}], \times \dots \times [a_{n},b_{n}] \times \mathbb{R} \times \dots \times \mathbb{R} \times \dots$ would be [[compact]], by [[closed subspaces of compact spaces are compact]]. But $\overline{B}$ is not [[compact]] (because $\mathbb{R}$ is not). ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```