Compact embedding

Let ${\displaystyle X}$ be a topological space, and let ${\displaystyle V}$ and ${\displaystyle W}$ be subsets of ${\displaystyle X}$. We say that ${\displaystyle V}$ is compactly embedded in ${\displaystyle W}$ if

• ${\displaystyle V\subseteq \operatorname {Cl} (V)\subseteq \operatorname {Int} (W)}$, where ${\displaystyle \operatorname {Cl} (V)}$ denotes the closure of ${\displaystyle V}$, and ${\displaystyle \operatorname {Int} (W)}$ denotes the interior of ${\displaystyle W}$; and
• ${\displaystyle \operatorname {Cl} (V)}$ is compact.

Equivalently, it states that there exists some compact set ${\displaystyle K}$, such that ${\displaystyle V\subseteq K\subseteq \operatorname {Int} (W)}$.

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be two normed vector spaces with norms ${\displaystyle \|\cdot \|_{X}}$ and ${\displaystyle \|\cdot \|_{Y}}$ respectively, and suppose that ${\displaystyle X\subseteq Y}$. We say that ${\displaystyle X}$ is compactly embedded in ${\displaystyle Y}$, if

• ${\displaystyle X}$ is continuously embedded in ${\displaystyle Y}$; i.e., there is a constant ${\displaystyle C}$ such that ${\displaystyle \|x\|_{Y}\leq C\|x\|_{X}}$ for all ${\displaystyle x}$ in ${\displaystyle X}$; and
• The embedding of ${\displaystyle X}$ into ${\displaystyle Y}$ is a compact operator: any bounded set in ${\displaystyle X}$ is totally bounded in ${\displaystyle Y}$, i.e. every sequence in such a bounded set has a subsequence that is Cauchy in the norm ${\displaystyle \|\cdot \|_{Y}}$.

If ${\displaystyle Y}$ is a Banach space, an equivalent definition is that the embedding operator (the identity) ${\displaystyle i\colon X\to Y}$ is a compact operator.