Total Subset

Last updated Nov 1, 2022

Let $(X, ||\cdot||)$ be a Normed Vector Space. We say $A \subset X$ is a Total Subset if $\text{cl}(\text{span}A) = X$. That is, $\text{span} A$ is Dense in $X$.

1. Since The Subspace Span is the Set of all Linear Combinations, this means we can get arbitrarily close to any $x \in X$ using some Linear Combination from $A$. Equivalent conditions density on metric spaces shows that this is equivalent to $\text{span}A$ being Dense, making $A$ a Total Subset.
2. Extending (1), we can get a Sequence $({x}{n}){n=1}^{\infty} \subset \text{span}A$ so that $||x_{n} - x|| \to 0$. Then, we can take the Telescoping Sum $$x_{1} = \sum\limits_{n=2}^{\infty} x_{n} - x_{n-1} = x.$$since each $x_{n} \in \text{span}A$ for $n \in \mathbb{N}$, we can break it into a (finite) Linear Combination of elements from $A$. Together, this means we can write $x$ as an Infinite Linear Combination of elements from $A$. We can go the other direction by noting each finite Linear Combination from our Infinite Linear Combination is in $\text{span}A$. This means for each $x \in X$, we have a Sequence in $\text{span}A$ that converges to $x$. Thus, by equivalent conditions on density on metric spaces, $\text{span}A$ is Dense and $A$ is a Total Subset.