Limit Points and Housekeeping (2.6-9)

Walter Rudin Principles of Mathematical Analysis, chapter 2, exercises 6-9:

6. Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed. Prove that $E'=\overline{E}'$. Generally, is it true that $E'=E''$?
7. Let $A_n$ be subsets of a metric space for $n∈ℕ$.
(a) If $B_n=∪_{i=0}^n A_i$ prove that $\overline{B_n}=∪_{i=0}^n \overline{A_i}$.
(b) If $B=∪A_n$ prove that $\overline{B}⊇∪\overline{A_n}$.
Show that this inclusion can be proper.
8. Is every point of every open set $E⊆ℝ^2$ a limit point of $E$? Answer the same question for closed sets in $ℝ^2$.
9. Let $E ^\circ$ denote the set of all interior points of a set $E$, called the interior of $E$.
(a) Prove that $E ^\circ$ is always open.
(b) Prove that $E$ is open iff $E ^\circ = E$.
(c) If $G ⊆ E$ and $G$ is open, show $G⊆E ^\circ$.
(d) Prove ${E^\circ}^c = \overline{E^c}$.
(e) Does $E ^\circ = \overline{E}^\circ$?
(f) Does $\overline{E}=\overline{E ^\circ}$?

Proof: Lemma 1: Let $∪A_i$ be a union of subsets of a metric space. Then $(∪A_i)'⊇∪A_i'$, and there is equality when the union is finite. Proof: The $⊇$ direction is clear since limit points of a subset are limit points of the whole set, so it remains to show $⊆$ when the union is finite. To wit, let $p$ be a limit point of $∪A_i$. For each $i$, let $r_i=\text{inf }\{r∈ℝ~|~A_i∩N_r(p)-\{p\}≠∅\}$. If $r_i > 0$ for all $i$ then since the indices are finite choose some $r$ such that $0 < r < \text{min}_i~r_i$ and we would have $N_r(p)∩(∪A_i)=\{p\}$ and $p$ is not a limit point, a contradiction. Hence $r_i=0$ for some $i$, therefore $p$ is a limit point of $A_i$.

6. We now show $E''⊆E'$ so that $E'$ is closed. We have in fact already seen this, since a limit of limit points is a limit of the original points, since they lie arbitrarily close to these limit points. Now, by the lemma we see $\overline{E}'=(E∪E')'=E'∪E''=E'$. Finally, let $E=(0,1)$. We see $E'=\{0,1\}$ and $E''=∅$ so that generally $E'$ need not equal $E''$.

7. (a) We see$$\overline{B_n}=B_n∪B_n'=(∪_{i=0}^n A_i)∪(∪_{i=0}^n A_i)'=$$$$(∪_{i=0}^n A_i)∪(∪_{i=0}^n A_i')=∪_{i=0}^n (A_i∪A_i')=∪_{i=0}^n \overline{A_i}$$with a $⊇$ containment between the appropriately modified terms three and four if the union is infinite. When $A_i=\{1/(i+1)\}$ for all $i$, we see $\overline{B}$ contains $0$, whereas $∪\overline{A_i}=∪A_i$ does not.

8. $ℝ^2$ contains no isolated points, and every point of an open set is interior, and since interior and non-isolated implies limit point, every point of an open set in $ℝ^2$ is a limit point. Clearly, $\{0\}$ is a closed set for which $0$ is not a limit point.

9. Lemma 2: $E^\circ = \overline{E^c}^c$. Proof: ($⊆$) Suppose $p∈E^\circ$. Then $p∈E$ so $p∉E^c$ and also $p∉{E^c}'$, hence $p∉\overline{E^c}$ so $p∈\overline{E^c}^c$. ($⊇$) Suppose $p∈\overline{E^c}^c$, so $p∉\overline{E^c}=E^c∪{E^c}'$ implying $p∈E$ and $p$ is not a limit point of $E^c$, i.e. $p$ is an interior point of $E$ and is contained in $E$.

(a) This follows from the lemma since $\overline{E^c}$ is closed.

(b) This will follow from (c).

(c) Assume $G⊆E$ is open and $G⊈E^\circ$. Then $G'=G∪E^\circ$ is an open set contained in $E$ with inclusions $E^\circ = \overline{E^c}^c ⊂ G' ⊆ E$. We then observe $E^c ⊆ G'^c ⊂ \overline{E^c}$ are inclusions with $G'^c$ closed containing $E$, yet $\overline{E^c}$ is supposed to be the smallest closed set containing $E^c$, a contradiction.

(d) This was shown in the lemma.

(e) Let $E=ℝ-\{0\}⊆ℝ^1$. Then $E^\circ=E$ yet $\overline{E}^\circ=ℝ^\circ=ℝ$. Hence, generally, $E ^\circ ≠ \overline{E}^\circ$.

(f) Let $E=\{0\}$. Then $\overline{E}=E$ and $\overline{E^\circ}=\overline{∅}=∅$. Hence, generally, $\overline{E}≠\overline{E ^\circ}$.$~\square$