# proper subspaces of a topological vector space have empty interior

Theorem - Let $V$ be a topological vector space^{}. Every proper subspace^{} $S\subset V$ has empty interior.

Proof : Let $S$ be a subspace of $V$. Suppose there is a non-empty open set $A\subseteq S$.

Fix a point ${a}_{0}\in A\subseteq S$. Since the vector sum operation is continuous, translations^{} of open sets are again open sets. In particular, the set $A-{a}_{0}:=\{x-{a}_{0}:x\in A\}$ is an open set of $V$ that contains the origin $0$.

As $S$ is a vector subspace and $A\subseteq S$, we see that the translation $A-{a}_{0}$ is still contained in $S$.

Since the scalar multiplication operation is continuous it follows easily that, for every $x\in V$, the function ${f}_{x}:\mathbb{K}\u27f6V$ given by

$${f}_{x}(\lambda )=\lambda x$$ |

is also continuous.

Consider now any vector $v\in V$. The set ${f}_{v}^{-1}(A-{a}_{0})$ is an open set that contains $0$. Thus, taking a value $\lambda \in {f}_{v}^{-1}(A-{a}_{0})$ we see that

$$\lambda v\in A-{a}_{0},$$ |

i.e. we can multiply $v$ by a sufficiently small $\lambda $ such that $\lambda v$ belongs to the open set $A-{a}_{0}$.

Since the set $A-{a}_{0}$ is contained in $S$, we see that $\lambda v\in S$, and therefore $v\in S$.

This proves that $V=S$, i.e. $S$ is not proper.

We conclude that if $S$ is proper then $S$ has empty interior. $\mathrm{\square}$

Title | proper subspaces of a topological vector space have empty interior |
---|---|

Canonical name | ProperSubspacesOfATopologicalVectorSpaceHaveEmptyInterior |

Date of creation | 2013-03-22 17:34:23 |

Last modified on | 2013-03-22 17:34:23 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 6 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 46A99 |