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Let $X$ be a normed linear space with norm $||\cdot||$ and $A \neq \emptyset$ is a linear subspace of $X$. Prove that $\bar{A}$ is also a linear subspace of $X$.

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What have you tried? –  Alex Becker Sep 25 '12 at 18:48
    
I m not able to visualize the additive and multiplicative closure of the new two points added to the subspace closure[A] –  nimesh.iitd Sep 25 '12 at 18:54
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Show that the closure $\overline Y$ of a subspace $Y$ of a normed space $X$ is again a vector subspace.

Proof: Suppose $x, y \in \overline Y$. There are three cases we must consider.

  1. If $x, y \in Y$ then it is immediate that $x + y \in Y$ since $Y$ is a subspace of $X$ and thus $x + y \in \overline Y$.
  2. If $x \in \overline Y \setminus Y$ and $y \in Y$ then by definition, given $\varepsilon > 0$ there exists $x_0 \in Y$ with $x_0 \neq x$ such that $d(x_0, x) = \|x_0 - x\| < \varepsilon$. Now observe that since $x_0 \neq x$ then $x_0 + y \neq x + y$ and $x_0 + y \in Y$. Now we have $d(x_0 + y, x + y) = \|x_0 + y - x - y\| = \|x_0 - x\| < \varepsilon$. Hence we can conclude that $x_0 + y$ is an accumulation point of $Y$. Hence $x + y \in \overline Y$.
  3. If $x, y \in \overline Y \setminus Y$ then by definition given $\varepsilon > 0$ there exists $x_0 \in Y$ with $x_0 \neq x$ such that $d(x_0, x) = \|x_0 - x\| < \varepsilon/2$. Similarly, there exists $y_0 \in Y$ with $y_0 \neq y$ such that $d(y_0, y) = \|y_0 - y\| < \varepsilon/2$. If $x_0 + y_0 = x + y$ choose some $x_0 \neq x_0' \in Y$ such that $d(x_0', x) = \|x_0' - x\| < \varepsilon/2$ which is true since there are infinitely many elements in $Y$ with that property. So suppose without loss of generality that $x_0 + y_0 \neq x + y$. It follows that $$d(x_0 + y_0, x + y) = \|x_0 + y_0 - x - y\| \leq \|x_0 - x\| + \|y_0 - y\| < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon.$$ Hence we have shown that $x + y$ is an accumulation point of $Y$ and hence $x + y \in \overline Y$.

Now suppose $x \in \overline Y$. Suppose $\alpha$ is a scalar. We must consider two cases.

  1. If $x \in Y$ then since $Y$ is a subspace of $X$ it follows immediately that $\alpha x \in Y$ and thus $\alpha x \in \overline Y$.
  2. If $x \in \overline Y \setminus Y$ then $x$ is an accumulation point of $Y$. By definition given $\varepsilon > 0$ there exists $x_0 \in Y$ with $x_0 \neq x$ such that $d(x_0, x) = \|x_0 - x\| < \varepsilon/(\vert \alpha \vert + 1)$. Now since $x_0 \neq x$ then $\alpha x_0 \neq \alpha x$ and observe that $$d(\alpha x_0, \alpha x) = \|\alpha x_0 - \alpha x \| = \vert \alpha \vert \|x_0 - x\| < \vert \alpha \vert \frac{\varepsilon}{\vert \alpha \vert +1} < \varepsilon.$$ Thus we have shown that $\alpha x$ is an accumulation point and we can conclude that $\alpha x \in \overline Y$.

Thus we can conclude that $\overline Y$ is a subspace of $X$.

Note: It is my understanding that the closure of a set is the set along with all its limit points or accumulation points. It is my understanding that it NOT implied that all the elements of a set are limit points of the set. Please correct me if I'm wrong here!

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It is sufficient to prove that $\alpha x + \beta y \in \overline{X}$ where $\alpha, \beta$ are in the underlying field $\mathbb{F}$ and $x, y \in \overline{X}$. We know that $0 \in \overline{X}$ since $X \subset \overline{X}$. Since $x, y \in \overline{X}$ there exist $x_{j}, y_{j} \in X$ such that $x_j \to x$ and $y_j \to y$. Since multiplication and addition are continuous $\alpha x_j + \beta y_j \to \alpha x + \beta y$. Therefore, $\alpha x + \beta y \in \overline{X}$

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thanks a lot !!! –  nimesh.iitd Sep 25 '12 at 19:04
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