# Compact embedding and dense embedding

1. Oct 31, 2013

### Tatianaoo

2. Oct 31, 2013

### R136a1

Could you first start by defining what compact embedding and dense embedding means. I have some natural guesses, but I would like to know for sure what you mean.

Second, what kind of relationships do you want to consider? As it stands, your question is a bit vague. Obviously, you will find an entire host of relationships, but can you make it a bit more specific?

3. Nov 1, 2013

### Tatianaoo

True. By compact embedding I mean the following: let $(X,||\cdot||),~(Y,||\cdot||)$ be Banacch spaces and $X\subseteq Y$. We say that $X$ is compactly embedded in $Y$ if the identity operator $i:X\rightarrow Y$ is compact.

Actually by dense embedding I mean usual density of one space in another (a subset $A$ of a topological space $X$ is called dense (in $X$) if every point $x\in X$ either belongs to $A$ or is a limit point of $A$).

My question is: if I know that $A$ is compactly embedded in $X$, what can I say about density of $A$ in $X$?

Thanks for help!

4. Nov 1, 2013

### economicsnerd

Unqualified, you can't say one way or another.

Consider the case in which $X$ is a finite-dimensional linear subspace of $(Y,||\cdot||_Y)$ (and endow $X$ with any norm $||\cdot||_X$). Then $X$ is certainly compactly embedded; indeed the inclusion is finite-rank. But $X$ is also guaranteed to be $||\cdot||_Y$-closed.

Summing up:
- If $X=Y$ is finite-dimensional, then $X$ is compactly embedded in $Y$ and dense in $Y$.
- If $X$ is finite-dimensional and $X\subsetneq Y$, then $X$ is compactly embedded in $Y$ and not dense in $Y$.

Last edited: Nov 1, 2013
5. Nov 1, 2013

### Tatianaoo

Thanks a lot for clarifying this! Maybe there is some extra condition (except being compactly embedded) that subspace needs to satisfy in order to be dense? Or somebody could suggest nice book about this subject?