# Some v basic stuff from Rudin’s Analysis

Section 2.18:

Definition: Let X be metric space. All points and sets mentioned before are understood to be elements and subsets of X.

(a) A neighbourhood of a point p is a set $N_{r}(p)$ consisting of all points q such that $d(p,q). The number r is called the radius of $N_{r}(p)$.

(b) A point p is a limit point of the set E if every neighbourhood of p contains a point $q \neq p$ such that $q \in E$.

(c) If $p \in E$ and p is not a limit point of E, then p is called an isolated point of E.

(d) E is closed if every limit point of E is a point of E.

(e) A point p is an interior point if there is a neighbourhood N of p such that $N \subset E$.

(f) E is open if every point of E is an interior point of E.

(g) The complement of E (denoted by $E^{c}$) is the set of all points $p \in X$ such that $p \notin E$.

(h) E is perfect if E is closed and if every point of E is a limit point of E.

(i) E is bounded if there is a real number M and a point $q \in X$ such that $d(p,q) for all points p in E.

(j) E is dense in X if every point of X is a limit point of E, or a point of E (or both).

Let us note that in $R^{1}$ neighbourhoods are segments, whereas in $R^{2}$ neighbourhoods are interiors of circles.

2.19 Theorem: Every neighbourhood is an open set.

2.20 Theorem If p is a limit point of a set E, then every neighbourhood of p contains infinitely many points of E.

2.20 Corollary A finite point set has no limit points

2.21 Examples Let us consider the following subsets of $R^{2}$.

2.21 a: The set of all complex z such that $|z|<1$ This is not closed, it is open, it is not perfect set but it is bounded

2.21b: The set of all integers z such that $|z| \leq 1$ This is closed set, it is not open set, it is a perfect set and it is also bounded.

2.21c: A finite set: Such a set is vacuously closed, it is not open, it is not perfect, it is bounded.

2.21d: The set of all integers. This is a closed set in $R^{2}$, this set is not open in $R^{2}$, this is not a perfect set in $R^{2}$, and it is not bounded

2.21e: The set consisting of the numbers $\frac{1}{n}$ where $n \in \mathcal{N}$. Let us note that this set E has a limit point, namely, $z=0$ but that no point of E is a limit point of E, we wish to stress the difference between having a limit point and containing one. What can we say about the basic topological properties of this set ? It is not closed in $R^{2}$, it is not open in $R^{2}$, it is not perfect in $R^{2}$ but it is bounded in $R^{2}$.

2.21f: The set of all complex numbers that is $R^{2}$: this is closed in $R^{2}$, this is also open in $R^{2}$, this is also perfect set in $R^{2}$, but this set is not bounded.

2.21g: the segment $(a,b)$. This set is not closed in $R^{2}$, it is not open if we regard it as an open set in $R^{2}$, whereas it is open in $R^{1}$, and it is also bounded in $R^{2}$.

NOTE: It is very enriching to compare all the above definitions with the following basic starting point of topology: A set S is said to be topologized if we can answer the following question: Given any real point p and any subset X we can answer the question: is p a limit point of X ? There are two extreme cases here: each point is a limit point….in this case there are simply too many limit points, this is a worthless topology; on the other extreme, there is the case in which no point is a limit point. This is called discrete topology. The very fact that this is endowed with a name means this is not as worthless as the other case !! Now this in view, compare all above definitions with this. Remember Rudin’s emphasis is analysis whereas here it is topology which is emphasised.

2.22 Theorem Let $\{ E_{\alpha}\}$ be an arbitrary (finite or infinite) collection of sets $E_{\alpha}$. Then $(\bigcup_{\alpha}E_{\alpha})^{c} = \bigcap_{\alpha}(E_{\alpha}^{c})$

2.23 A set E is open if and only if its complement is closed.

2.23 Corollary: A set F is closed if and only if its complement is open.

2.24 Theorem:

2.24a: For any collection $\{ G_{\alpha}\}$ of open sets, $\bigcup_{\alpha}G_{\alpha}$ is open.

2.24b: For any collection $\{ F_{\alpha}\}$ of closed sets, $\bigcap_{\alpha}F_{\alpha}$ is closed.

2.24c: For any finite collection $G_{1}, G_{2}, \ldots, G_{n}$ of open sets $\bigcap_{i=1}^{n}G_{i}$ is open.

2.24d: For any finite collection $F_{1}, F_{2}, \ldots, F_{n}$ of closed sets, $\bigcup_{i=1}^{n}F_{i}$ is closed.

2.25 Example: In parts (c) and (d) of the preceding theorem, the finitiness of the collections is essential. For, let $G_{n}$ be the segment $(-\frac{1}{n}, \frac{1}{n})$ where $n=1,2,3,\ldots$. Then, $G_{n}$ is an open subset of $R^{1}$. Put $G=\bigcap_{i=1}^{\infty}G_{n}$. Then G consists of a single point (namely, x=0) and is, therefore not an open subset of $R^{1}$.

Thus, the intersection of an infinite collection of open sets need not be open. Similarly, the union of an infinite collection of closed sets need not be closed.

2.26 Definition: If X is a metric space, if $E \subset X$ and if $E^{'}$ denotes the set of all limit points of E in X, then the closure of E is the set $\overline{E}=E \bigcup E^{'}$

(Remark: The following few theorems, in my theorem, give an approach to solving problems in analysis: )

(Remark: I have tried to fill the missing gaps in Prof Rudin’s terse, distilled proofs!)

2.27 Theorem: If X is a metric space and $E \subset X$, then prove that

2.27a: $\overline{E}$ is closed

2.27b: $E=\overline{E}$ if and only if E is closed.

2.27c: $\overline{E} \subset F$ for every closed set $F \subset X$ such that $E \subset F$.

By (a) and (c), $\overline{E}$ is the smallest closed subset of X that contains E.

Proof:

2.27a: If $p \in X$ and $p \notin \overline{E}$, then p is neither a point of E nor a limit point of E. Hence, p has a neighbourhood which does not intersect E. The complement of $\overline{E}$ is therefore open. Hence, $\overline{E}$ is closed.

2.27b: If $E = \overline{E}$, from part a, we conclude that E is closed. Conversely, if E is closed, then $E^{'} \subset E$ and by definition of closed and 2.26 theorem above, we get that $\overline{E} =E$.

2.27c: Given that F is a closed set, and that $F \supset E$, then $F \supset F^{'}$ hence, $F \supset E^{'}$. Thus, $F \supset E$.

2.28 Theorem : Let E be a nonempty set of real numbers which is bounded above. Let $y = \sup {E}$. Then prove that $y \in \overline{E}$. Hence, $y \in E$ if E is closed.

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Remark: The author Prof Rudin wants us to pause now and compare above with the examples in Sec 1.9. I reproduce Sec 1.8 and Sec 1.9 below for easy reference and understanding:

Sec 1.8: Definition: Suppose S is an ordered set, with $E \subset S$, and E is bounded above. Suppose there exists an $\alpha \in S$ with the following properties:

(i) $\alpha$ is an upper bound of E.

(ii) If $\gamma < \alpha$ then $\gamma$ is not an upper bound of E. Then $\alpha$ is called the least upper bound of E (that there is at most one such $\alpha$ is clear from (ii)) or the supremum of E, and we write:

$\alpha = \sup{E}$.

The greatest lower bound or infimum of a set E which is bounded below is defined in the same manner. The statement $\alpha = \inf{E}$ means that $\alpha$ is a lower bound of E and that no $\beta$ with $\beta > \alpha$ is a lower bound.

Sec 1.9 Examples:

Sec 1.9a example: Let $llatex p^{2}=2$. Let A be the set of all positive rationals p such that $p^{2}<2$ and let B consist of all positive rationals p such that $p^{2}=2$. A and B are subsets of the ordered set Q. The set A is bounded above. In fact, the upper bounds of A are exactly the members of B. Since B contains no smallest member, A has no least upper bound in Q.

Similarly, B is bounded below. The set of all lower bounds of B consists of A and of all $r \in Q$ with $r \leq 0$. Since A has no largest member, B has no greatest lower bound in Q.

Sec 1.9b example: If $\alpha = \sup {E}$ exists, then $\alpha$ may or may not be a member of E. For instance, let $E_{1}$ be the set of all $r \in Q$ with $r <0$. Let $E_{2}$ be the set of all $r \in Q$ with $r \leq 0$. Then

$\sup {E_{1}}= \sup{E_{2}}=0$

and $0 \notin E_{1}$ and $0 \in E_{2}$.

Sec 1.9c example: Let E consist of all numbers $\frac{1}{n}$ where $n=1,2,3,\ldots$. Then $\sup {E}=1$, which is in E and inf E = 0, which is not in E.

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Now, we come back to our current discussion:

2.28 Theorem: Let E be a nonempty set of real numbers which is bounded above. Let $y = \sup {E}$. Then prove that $y \in E$ if E is closed.

Proof: If $y \in E$, then $y \in \overline{E}$. Assume $y \notin E$. For every $h >0$, there exists then a point $x \in E$ such that $y-h, for otherwise $y-h$ would be an upper bound of E. Thus, y is a limit point of E. Hence, $y \in \overline{E}$

2.29 Remark Suppose $E \subset Y \subset X$ where X is a metric space. To say that E is an open subset of X means that to each point $p \in E$, there is associated a positive number r such that the conditions $d(p,q), $q \in X$ imply that $q \in E$. But we have already observed (Sec 2.16) that Y is also a metric space, so that our definitions may equally well be made within Y. To be quite explicit, let us say that E is open relative to Y if to each $p \in E$ there is associated an $r>0$ such that $q \in E$ whenever $d(p,q) and $q \in Y$. Example 2.21g showed that a set may be open relative to Y without being an open subset of X. However, there is a simple relation between these concepts, which we now state.

2.30 Theorem Suppose $Y \supset X$. A subset E of Y is open relative to Y if and only if $E =Y \bigcap G$ for some open subset G of X.

Proof:

Suppose E is open relative to Y. To each $p \in E$, there is a positive number $r_{p}$ such that the conditions $d(p,q) and $q \in Y$ imply that $q \in E$. Let $V_{q}$ be the set of all $q \in X$ such that $d(p,q) and define

$G = \bigcup_{p \in E}V_{p}$

Thus G is an open subset of X, by theorems 2.19 and 2.24.

Since $p \in V_{p}$ for all $p \in E$, it is clear that $E \subset G \bigcap Y$.

By our choice of $V_{p}$, we have $V_{p} \bigcap Y \subset E$ for every $p \in E$, so that $G \bigcap Y \subset E$. Thus, $E = G \bigcap Y$, and one half of the theorem is proved.

Conversely, if G is open in X and $E = G \bigcap Y$, every $p \in E$ has a neighbourhood $V_{p} \subset G$. Then $V_{p} \bigcap Y \subset E$, so that E is open relative to Y.

Cheers,

Nalin Pithwa

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