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Intuition of: Every proper subspace of a normed vector space has empty interior?

Since there's a proof for it, then it's true, but I have troubles understanding the intuition of a "filled space" having every subspace with empty interiors.

Does it mean that the structure of normed spaces is somehow "discrete"? That it has holes in it?

Community
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mavavilj
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1 Answers1

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The intuition here is that if $V$ is a normed vector space, and $B$ is some open ball around the origin of $V$, then $B$ "contains every direction" in $V$. Indeed, if $v \in V$ is any vector, then I can always pick $\epsilon$ small enough such that $\epsilon v \in B$. So if $U \subseteq V$ is a proper subspace, then every open ball $B$ around the origin has points not belonging to $U$, so cannot be an interior point.

Imagine any proper subspace in $\mathbb{R}^3$, so a point, a line, or a plane. It is impossible to put a solid ball, however small, properly inside that subspace. There are no holes in the subspaces, it's just that the ball you are using still comes from the ambient space $\mathbb{R}^3$.

Joppy
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  • So it's more related to the concept of density of sets? – mavavilj Apr 28 '18 at 16:11
  • No, it's the fact that if $U \subseteq V$ is a linear subspace, by default we just treat this as a subset $U \subseteq V$ topologically. This means we are always using open balls in the ambient space. There is also a fine and well-defined way to restrict the norm to the subspace $V$, in which case it will have full interior (since we have now forgotten about the ambient space entirely). – Joppy Apr 28 '18 at 16:15