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Nov 17, 2023 · A subset $A$ of a metric space $X$ is relatively compact if and only if every sequence of points in $A$ has a cluster point in $X$. A space is compact if it is relatively compact in itself.
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Is $a$ relatively compact in $X$?
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What is a relatively compact subspace?
In mathematics, a relatively compact subspace (or relatively compact subset, or precompact subset) Y of a topological space X is a subset whose closure is compact.
A set of the form $C \cap E$ where $C$ is compact in $X$ is relatively compact in $E$ (when $X$ is Hausdorff), as $\operatorname{cl}(C \cap E) \subseteq C$ and so is a closed subset of a compact set, hence compact.
Sep 5, 2021 · A subset \(A\) of \(\mathbb{R}\) is compact if and only if it is closed and bounded. Proof. Suppose \(A\) is a compact subset of \(\mathbb{R}\). Let us first show that \(A\) is bounded. Suppose, by contradiction, that \(A\) is not bounded. Then for every \(n \in \mathbb{N}\), there exists \(a_{n} \in A\) such that \[\left|a_{n}\right| \geq n.\]
May 19, 2023 · The set $A_n = [-n,n] \setminus \mathbb Q$ in topology.pi-base.org/spaces/S000059 is shown to be relatively compact, and no compact set can contain any non-degenerate interval.
A subset of the Banach space of real-valued continuous functions on a compact Hausdorff space is relatively compact if and only if it is equicontinuous and pointwise bounded (Arzelà–Ascoli theorem).
Feb 8, 2018 · For clarity, a set $E$ is said to be relatively compact if its closure $\overline{E}$ is compact. Attempt at a proof: Assume $E$ is totally bounded. Then by a corollary, we know that every sequence in $E$ has a Cauchy subsequence.
