Closure of Subspace of Normed Vector Space is Subspace

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Theorem

Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.

Let $Y \subseteq X$ be a subspace of $X$.

Let $Y^-$ be the closure of $Y$.


Then $Y^- \subseteq X$ is also a subspace of $X$.


Proof

Suppose $y \in Y^-$.

Then there is a sequence $\displaystyle \sequence {y_n}_{n \mathop \in \N} \in Y$ which converges to $y$.

Suppose $y \in Y$ and $y$ is a limit point.

Then one can define a constant sequence $\sequence {y_n}_{n \mathop \in \N} = y$.


Closed under restriction of vector addition

Let $x, y \in Y^-$.

Let $\sequence {x_n}_{n \mathop \in \N}, \sequence {y_n}_{n \mathop \in \N}$ be sequences in $Y$.

Suppose:

$\displaystyle \lim_{n \mathop \to \infty} x_n = x$
$\displaystyle \lim_{n \mathop \to \infty} y_n = y$

Since $Y$ is a subspace:

$\forall n \in \N : x_n + y_n \in Y \subseteq Y^-$

Furthermore:

$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = x + y$

$Y^-$ is closed and contains its limit points.

Hence, $x + y \in Y^-$.

$\Box$

Closed under restriction of scalar multiplication

Let $\alpha \in \Bbb K$, where $\Bbb K$ is a field.

Let $y \in Y^-$.

Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $Y$.

Suppose:

$\displaystyle \lim_{n \mathop \to \infty} y_n = y$

Since $Y$ is a subspace:

$\forall n \in \N : \alpha \cdot y_n \in Y \subseteq Y^-$

Furthermore:

$\displaystyle \lim_{n \mathop \to \infty} \alpha \cdot y_n = \alpha \cdot y$.

$Y^-$ is closed and contains its limit points.

Hence:

$\alpha \cdot y \in Y^-$

$\Box$

Nonemptiness

$Y$ is a subspace.

Hence:

$0 \in Y \subseteq Y^-$

Thus $Y^-$ contains at least one element and is non-empty.


$\Box$

Hence, $Y^-$ is closed subspace

$\blacksquare$


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