If $A$ is a subset of $(V,\| \cdot \|)$, where $V$ is a vector space. Let $\overline{A}$ denote closure. Show $\overline{x + A} = x + \overline{A}$

Question

Let $A$ a subset of $(V,\| \cdot \|)$, where $V$ is a vector space. Let $\overline{A}$ denote closure of $A$. Show that if $x \in V$ then $\overline{x + A} = x + \overline{A}$.

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Attempt:

$(\Rightarrow)$ Consider $y \in \overline{x + A} = (x + A) \cup (x + A)'$ where $(x + A)'$ is the set of limit points.
$$ \Rightarrow \quad y \in x + A \quad \textrm{or} \quad y \in (x + A)'$$

Scenario 1: $y \in x + \overline{A}$ (in which case we are done).

Scenario 2: $ y \in x'$ and $y \in A'$.

Continuing with Scenario 2: Since $y \in x'$ but $x'$ is just an element which means $y = x = \overline{x}$. Since $y \in A'$ this means $y \in \overline{A}$. Therefore $y \in x + \overline{A}$.

$(\Leftarrow)$ Let $y \in x + \overline{A}$.

Then $y = x$ and $y \in A \cup A'$

This means: $y \in x + A$ and $y \in x + A'$.

Scenario 1: $y \in x + A$. If this is the case we are done because $x + A \subset \overline{x + A}$.

Scenario 2: $y \in x + A'$. This means $y = x$ and $y \in A'$. Which means $y\ = x = \overline{x}$ and $y \in A'$ means $y \in \overline{A}$. Which implies $y \in \overline{x + A}$.

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Comment: I may be overthinking thing a bit, but I don't really think this is the right way to answer this problem. I never used the notion of norm given to me and I feel that does have a key part to play. Feedback on my approach ?

Answer 1

It is easier to show that $y\in \overline{x+A}$ if and only if $y-x\in\overline{A}.$

If $y-x\in\overline{A}$ then for any $\epsilon>0$ there is an $a\in A$ such that $\|y-x-a\|<\epsilon.$ But then letting $a'=x+a\in x+A,$ we have for any $\epsilon>0$ an $a'\in x+A$ such that $\|y-a'\|<\epsilon,$ so $y\in \overline{x+A}.$

On the other hand, if $y\in \overline{x+A}$ then for any $\epsilon>0$ there exists $a'\in x+A$ with $\|y-a'\|<\epsilon.$ then letting $a=a'-x\in A$ we have, for any $\epsilon>0$ an $a\in A$ such that $\|y-x-a\|=\|y-a'\|<\epsilon.$ So $y-x\in \overline{A}.$

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From the comments, there is some confusion, perhaps because the readers do not know this lemma:

Lemma: If $(X,d)$ is a metric space and $A\subseteq X,$ then we have that $x\in\overline A$ if and only if for all $\epsilon>0$ there is an $a\in A$ such that $d(x,a)<\epsilon.$

If $x\in A,$ and any $\epsilon>0,$ you can choose $a=x\in A$ and get $d(x,a)=0<\epsilon.$

If $x\in A',$ then you can use the definition of limit point to prove this.

On the other hand, we need to show that if $x$ satisfies the $\epsilon$ statement for all $\epsilon>0$ then $x\in\overline{A}=A\cup A'.$

If $x\in A$ then $x\in\overline{A},$ so we only need to check that the cases where $x\not\in A,$ then show the $\epsilon$ statement implies $x\in A'.$

Answer 2

Here's another approach that you could consider.

Let $x$ be a fixed element of $V$, the function $f:V\to V$ defined by $$ f(y) := y+x $$ is continuous and has a continuous inverse $f^{-1}(y) = y-x$. This means that $f$ is a homeomorphism so $$ \overline{f(A)} = f(\overline A). $$ Unpacking the notation gives $\overline{x + A} = x + \overline{A}$ as you want.

Edit: I should also have mentioned that we used the fact that $h(\overline A) \subset \overline{h(A)}$ if $h$ is continuous to get the conclusion (apply this fact to $f$ and $f^{-1}$).

Answer 3

It is clear that $x+A=x+A$, so it suffices to check that $(x+A)'=x+A'$ to prove that $\overline{x+A}=x+\overline{A}$.

Consider a convergent sequence $(a_{n})\in A$. We have the limit $a\in A'$ and $x+a\in x+A'$, so $\|x+a_{n}-x-a\|\rightarrow 0$. This implies that $x+a$ is the limit of a sequence $(x+a_{n})$ in $(x+A)$. So $x+A'\subseteq(x+A)'$.

Now, consider a sequence $(x+a_{n})$ with limit $(x+a)\in(x+A)'$. We need to show that $a\in A'$. We have $\|a_{n}-a\|=\|x-a_{n}-x-a\|\rightarrow 0$. Therefore $a\in A'$ and so $x+a\in x+A'$. Hence $(x+A)'=x+A'$ and our conclusion follows.