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Viewing Version
2
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'closure of a vector subspace is a vector subspace'
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| Title of object: |
closure of a vector subspace is a vector subspace |
| Canonical Name: |
ClosureOfAVectorSubspaceIsAVectorSubspace2 |
| Type: |
Theorem |
| Created on: |
2005-02-04 13:20:26 |
| Modified on: |
2005-02-04 13:26:32 |
| Classification: |
msc:15A03, msc:46B99, msc:54A05 |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage{mathrsfs}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\newcommand{\sR}[0]{\mathbb{R}}
\newcommand{\sC}[0]{\mathbb{C}}
\newcommand{\sN}[0]{\mathbb{N}}
\newcommand{\sZ}[0]{\mathbb{Z}}
\usepackage{bbm}
\newcommand{\Z}{\mathbbmss{Z}}
\newcommand{\C}{\mathbbmss{C}}
\newcommand{\R}{\mathbbmss{R}}
\newcommand{\Q}{\mathbbmss{Q}}
\newcommand*{\norm}[1]{\lVert #1 \rVert}
\newcommand*{\abs}[1]{| #1 |}
\newtheorem{thm}{Theorem}
\newtheorem{defn}{Definition}
\newtheorem{prop}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{cor}{Corollary} |
Content:
\begin{thm} In a topological vector space
the \PMlinkname{closure}{Closure} of a vector subspace is a vector subspace.
\end{thm}
\begin{proof}
Let $X$ be the topological vector space over $\mathbbmss{F}$ where
$\mathbbmss{F}$ is either $\R$ or $\C$, and let $V$ be a vector subspace
in $X$.
To prove that the closure of $V$, $\overline{V}$,
is a vector subspace of $X$, it suffices
to prove that
$$
\lambda x + \mu y \in \overline{V}
$$
whenever $\lambda,\mu \in \mathbbmss{F}$ and $x,y\in \overline{V}$.
Suppose $\lambda,\mu,x,y$ are as above. Then
there are sequences $(x_i)$, $(y_i)$ in $V$ converging to
$x,y$, respectively.
In a topological vector space, addition and multiplication are continuous
operations. Thus
\begin{eqnarray*}
\lim_{i\to \infty} (\lambda x_i + \mu y_i) &=& \lim_{i\to \infty} \lambda x_i + \lim_{i\to \infty}\mu y_i \\
&=& \lambda \lim_{i\to \infty} x_i + \mu \lim_{i\to \infty} y_i \\
&=& \lambda x + \mu y,
\end{eqnarray*}
so $\lambda x_i + \mu y_i$ converges to $\lambda x + \mu y$.
We have proven that $\lambda x + \mu y \in \overline{V}$, so
$\overline{V}$ is a vector subspace.
\end{proof} |
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