hemicompact space

A topological space $(X,\\tau)$ is called a hemicompact space if there is an admissible sequence in $X$ , i.e. there is a sequence of compact sets $(K_{n})_{n\\in\\mathbb{N}}$ in $X$ such that for every $K\\subset X$ compact there is an $n\\in\\mathbb{N}$ with $K\\subset K_{n}$ .

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The above conditions imply that if $X$ is hemicompact with admissible sequence $(K_{n})_{n\\in\\mathbb{N}}$ then $X=\\bigcup_{n\\in\\mathbb{N}}K_{n}$ because every point of $X$ is compact and lies in one of the $K_{n}$ .
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A hemicompact space is clearly $\\sigma$ -compact. The converse is false in general. This follows from the fact that a first countable hemicompact space is locally compact (see below). Consider the set of rational numbers $\\mathbb{Q}$ with the induced euclidean topology. $\\mathbb{Q}$ is $\\sigma$ -compact but not hemicompact. Since $\\mathbb{Q}$ satisfies the first axiom of countability it can’t be hemicompact as this would imply local compactness.
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Not every locally compact space (like $\\mathbb{R}$ ) is hemicompact. Take for example an uncountable discrete space. If we assume in addition $\\sigma$ -compactness we obtain a hemicompact space (see below).

Proposition. Let $(X,\\tau)$ be a first countable hemicompact space. Then $X$ is locally compact.

Proof.

Let $\\cdots\\subset K_{n}\\subset K_{n+1}\\subset\\cdots$ be an admissible sequence of $X$ .Assume for contradiction that there is an $x\\in X$ without compact neighborhood. Let $U_{n}\\supset U_{n+1}\\supset\\cdots$ be a countable basis for the neighbourhoods of $x$ . For every $n\\in\\mathbb{N}$ choose a point $x_{n}\\in U_{n}\\setminus K_{n}$ . The set $K:=\\{x_{n}:n\\in\\mathbb{N}\\}\\cup\\{x\\}$ is compact but there is no $n\\in\\mathbb{N}$ with $K\\subset K_{n}$ . We have a contradiction.∎

Proposition. Let $(X,\\tau)$ be a locally compact and $\\sigma$ -compact space. Then $X$ is hemicompact.

Proof.

By local compactness we choose a cover $X\\subset\\bigcup_{i\\in I}U_{i}$ of open sets with compact closure (take a compact neighborhood of every point). By $\\sigma$ -compactness there is a sequence $(K_{n})_{n\\in\\mathbb{N}}$ of compacts such that $X=\\bigcup_{n\\in\\mathbb{N}}K_{n}$ . To each $K_{n}$ there is a finite subfamily of $(U_{i})_{i\\in I}$ which covers $K_{n}$ .Denote the union of this finite family by $U_{n}$ for each $n\\in\\mathbb{N}$ . Set $\\tilde{K}_{n}:=\\overline{\\bigcup_{k=1}^{n}U_{k}}$ . Then $(\\tilde{K}_{n})_{n\\in\\mathbb{N}}$ is a sequence of compacts. Let $K\\subset X$ be compact then there is a finite subfamily of $(U_{i})_{i\\in I}$ covering $K$ . Therefore $K\\subset K_{n}$ for some $n\\in\\mathbb{N}$ .∎

单词	HemicompactSpace
释义	hemicompact space A topological space $(X,\\tau)$ is called a hemicompact space if there is an admissible sequence in $X$ , i.e. there is a sequence of compact sets $(K_{n})_{n\\in\\mathbb{N}}$ in $X$ such that for every $K\\subset X$ compact there is an $n\\in\\mathbb{N}$ with $K\\subset K_{n}$ . • The above conditions imply that if $X$ is hemicompact with admissible sequence $(K_{n})_{n\\in\\mathbb{N}}$ then $X=\\bigcup_{n\\in\\mathbb{N}}K_{n}$ because every point of $X$ is compact and lies in one of the $K_{n}$ . • A hemicompact space is clearly $\\sigma$ -compact. The converse is false in general. This follows from the fact that a first countable hemicompact space is locally compact (see below). Consider the set of rational numbers $\\mathbb{Q}$ with the induced euclidean topology. $\\mathbb{Q}$ is $\\sigma$ -compact but not hemicompact. Since $\\mathbb{Q}$ satisfies the first axiom of countability it can’t be hemicompact as this would imply local compactness. • Not every locally compact space (like $\\mathbb{R}$ ) is hemicompact. Take for example an uncountable discrete space. If we assume in addition $\\sigma$ -compactness we obtain a hemicompact space (see below). Proposition. Let $(X,\\tau)$ be a first countable hemicompact space. Then $X$ is locally compact. Proof. Let $\\cdots\\subset K_{n}\\subset K_{n+1}\\subset\\cdots$ be an admissible sequence of $X$ .Assume for contradiction that there is an $x\\in X$ without compact neighborhood. Let $U_{n}\\supset U_{n+1}\\supset\\cdots$ be a countable basis for the neighbourhoods of $x$ . For every $n\\in\\mathbb{N}$ choose a point $x_{n}\\in U_{n}\\setminus K_{n}$ . The set $K:=\\{x_{n}:n\\in\\mathbb{N}\\}\\cup\\{x\\}$ is compact but there is no $n\\in\\mathbb{N}$ with $K\\subset K_{n}$ . We have a contradiction.∎ Proposition. Let $(X,\\tau)$ be a locally compact and $\\sigma$ -compact space. Then $X$ is hemicompact. Proof. By local compactness we choose a cover $X\\subset\\bigcup_{i\\in I}U_{i}$ of open sets with compact closure (take a compact neighborhood of every point). By $\\sigma$ -compactness there is a sequence $(K_{n})_{n\\in\\mathbb{N}}$ of compacts such that $X=\\bigcup_{n\\in\\mathbb{N}}K_{n}$ . To each $K_{n}$ there is a finite subfamily of $(U_{i})_{i\\in I}$ which covers $K_{n}$ .Denote the union of this finite family by $U_{n}$ for each $n\\in\\mathbb{N}$ . Set $\\tilde{K}_{n}:=\\overline{\\bigcup_{k=1}^{n}U_{k}}$ . Then $(\\tilde{K}_{n})_{n\\in\\mathbb{N}}$ is a sequence of compacts. Let $K\\subset X$ be compact then there is a finite subfamily of $(U_{i})_{i\\in I}$ covering $K$ . Therefore $K\\subset K_{n}$ for some $n\\in\\mathbb{N}$ .∎
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