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Show that if $A \subset B$ , then $\overline{A} \subset \overline{B}.$

$\overline{E}$ = The set of adherent points $x \in \textbf{R}$ such that $ \ \forall \epsilon>0 ,\ (x-\epsilon, x+ \epsilon) \cap E \neq \emptyset. $

Suppose that $ A \subset B$ is true then: $$\forall x \in A \implies x \in B.$$ Let's take a point $x_1$ in A such that $$\forall \epsilon>0 \ , \ (x_1 - \epsilon , x_1+\epsilon ) \cap A \neq \emptyset.$$ $$\iff x_1 \in \overline{A}.$$

Since $x_1$ is in $A$ , it follows that $x_1$ is in B because $A$ is a subset of B and also :

Im not sure if it's correct from here: 

$$(x_1-\epsilon ,x_1+\epsilon)\cap B \neq \emptyset. $$

We conclude that for all $x \in\overline{A}$, then $x \in \overline{B} \implies \overline{A} \subset \overline{B}.$

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  • $\begingroup$ "Adherent points", "Adherent set"? Whose bright idea was this terminology instead of the well known "accumulation points" and "closure"? $\endgroup$ – Paul Sinclair Jun 8 '16 at 22:21
  • $\begingroup$ Before someone else points it out, I must admit that accumulation points are not exactly the same thing, since isolated elements of the set are adherent points, but are not accumulation points, since an accumulation point must have points of the set other than itself in every neighborhood. However, the "adherent set" is exactly the closure. $\endgroup$ – Paul Sinclair Jun 8 '16 at 23:07
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You are misunderstanding the definition.

Let's take a point $x_1$ in A such that $$\forall \epsilon>0 \ , \ (x_1 - \epsilon , x_1+\epsilon ) \cap A \neq \emptyset.$$ $$\iff x_1 \in \overline{A}.$$

You don't start with $x_1 \in A$. The definition is for arbitrary $x \in \Bbb R$. If you restrict it to elements of $A$, then you would have $\overline A = A$ for any set $A$, and the concept would be useless.

You want to show that $\overline A \subset \overline B$. That is, every element of $\overline A$ is also in $\overline B$. So, start with

  • Let $x \in \overline A$

What does that mean?

  • $\forall \epsilon > 0, (x - \epsilon, x + \epsilon) \cap A \ne \emptyset$

I.e.,

  • $\exists\: x_1 \in (x - \epsilon, x + \epsilon)$ with $x_1 \in A$.

But

  • $A \subset B$, so $x_1 \in B$.

Therefore

  • $x_1 \in (x - \epsilon, x + \epsilon)$ and $x_1 \in B$.

I.e.,

  • $\forall \epsilon > 0, (x - \epsilon, x + \epsilon) \cap B \ne \emptyset$

And so

  • $x \in \overline B$.

Since this holds for arbitrary $x$,

  • $\overline A \subset \overline B$.
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  • $\begingroup$ Im confused now.. does an adherent point need to be in the set A? $\endgroup$ – Elina Jun 8 '16 at 23:50
  • $\begingroup$ @Neophyte No, it doesn't need to. $\endgroup$ – egreg Jun 8 '16 at 23:55
  • $\begingroup$ so what if the adherent set empty?.......I think by definition an empty set is a subset for any set.... not sure.. $\endgroup$ – Elina Jun 8 '16 at 23:58
  • $\begingroup$ thx guys, I understand now :) $\endgroup$ – Elina Jun 8 '16 at 23:59
  • $\begingroup$ Just to be sure (or in case someone else has the same question): Every point in $A$ is automatically an adherent point: For every $\epsilon > 0, x\in (x - \epsilon, x + \epsilon)$, so if $x \in A$, then $(x - \epsilon, x + \epsilon) \cap A \ne \emptyset$ contains $x$, and so is not empty. But there can also be adherent points not in $A$. For example, if $A = (0,1)$, then both $0$ and $1$ are adherent points even though they are not in $A$. Because the set of adherent points includes the set $A$, the only way for it to be empty is if $A = \emptyset$. $\endgroup$ – Paul Sinclair Jun 9 '16 at 1:19
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The closure of X is the intersection of the closed sets larger than X. A being smaller has more larger closed sets and thus a smaller intersection, where large and small are in terms of the subset relation.

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  • $\begingroup$ It seems apparent that Neophyte is just beginning to learn about the topology of the real line (unfortunately with a non-standard terminology). So it is unlikely that he or she would be familiar with these facts. $\endgroup$ – Paul Sinclair Jun 9 '16 at 1:29
  • $\begingroup$ @PaulSinclair My answer was not meant to be the best answer, or in accordance with "pedagogy". If he has some intellectual vigor however he might be able to contemplate on my proof and gain much more than from chasing elements. $\endgroup$ – Jacob Wakem Jun 9 '16 at 1:53

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