# VI Convergence. Open and Closed Sets.

Reference: Introductory Real Analysis by Kolmogorov and Fomin.

Reference: Introduction to Analysis by Rosenlicht.

Reference: Analysis by Walter Rudin.

Sec 6.1 Closure of a set. Limit points.

By the open sphere (or open ball) $S(x_{0},r)$ in a metric space R we mean the set of points $x \in R$ satisfying the inequality $\rho(x_{0},r)

($\rho$ is the metric of R.)

Note: Any confusion between “sphere” meant in the sense of spherical surface and “sphere” meant in the sense of a solid sphere (or ball) will always be avoided by judicious use of the adjectives “open” or “closed.”

The fixed point $x_{0}$ is called the center of the sphere and the number r is called its radius. By the closed sphere (or closed ball) $S[x_{0},r]$ with center $x_{0}$ and radius r we mean the set of points $x \in R$ satisfying the inequality

$\rho(x_{0},x) \leq r$

An open sphere of radius $\epsilon$ with center $x_{0}$ will also be called an $\epsilon$neighbourhood of $x_{0}$ denoted by $O_{\epsilon}(x_{0})$.

A point $x \in R$ is called a contact point of a set $M \subset R$ if every neighbourhood of x contains at least one point of M. The set of all contact points of a set M is denoted by $[M]$ and is called the closure of M. Obviously, $M \subset [M]$, since every point of M is a contact point of M. By the closure operator in a metric space R, we mean the mapping of R carrying each set $M \subset R$ into its closure $[M]$.

Theorem 1:

The closure operator has the following properties:

1. If $M \subset N$, then $[M] \subset [N]$
2. $[[M]] = [M]$
3. $[M \bigcup N] = [M] \bigcup [N]$
4. $[\phi] = \phi$.

Proof I:

Property I is obvious.

Proof of property 2:

Let $x \in [[M]]$. Then, any given neighbourhood $O_{\epsilon}(x)$ contains a point $x_{1} \in [M]$. Consider the sphere $O_{\epsilon_{1}}(x_{1})$ of radius

$\epsilon_{1} = \epsilon - \rho(x,x_{1})$

PS: It helps to draw an illustrative diagram to understand this proof.

Clearly, $O_{\epsilon_{1}}(x_{1})$ is contained in $O_{\epsilon}(x)$. In fact, if $z \in O_{\epsilon_{1}}(x_{1})$ then $\rho(z,x_{1}) < \epsilon_{1}$ and hence, since $\rho(x,x_{1}) = \epsilon - \epsilon_{1}$, it follows from the triangle inequality that

$\rho(z,x) < \epsilon_{1} + (\epsilon - \epsilon_{1}) = \epsilon$,

That is, $z \in O_{\epsilon}(x)$. Since $x_{1} \in [M]$, there is a point $x_{2} \in M$ in $O_{\epsilon_{1}}(x)$. But then $x_{2} \in O_{\epsilon}(x)$ and hence, $x \in [M]$, since $O_{\epsilon}(x)$ is an arbitrary neighbourhood of x. Therefore, $[[M]] \subset [M]$. But, obviously $[M] \subset [[M]]$ and hence, $[[M]] = [M]$, as required.

Proof of property 3:

Let $x \in [M \bigcup N]$ and suppose $x \notin [M] \bigcup [N]$. Then, $x \notin [M]$ and $x \notin [N]$. But then there exist neighbourhoods $O_{\epsilon_{1}}(x)$ and $O_{\epsilon_{2}}(x)$ such that $O_{\epsilon_{1}}(x)$ contains no points of M while $O_{\epsilon_{1}}(x)$ contains no points of N. It follows that the neighbourhood $O_{\epsilon}(x)$, where $\epsilon=\min (\epsilon_{1}, \epsilon_{2})$, contains no points of either M or N, and hence no points of $M \bigcup N$, contrary to the assumption that $x \in [M \bigcup N]$. Therefore, $x \in [M] \bigcup [N]$, and hence,

$[M \bigcup N] = [M] \bigcup [N]$…call this I

since x is an arbitrary point of $[M \bigcup N]$. On the other hand, since $M \subset M \bigcup N$ and $N \subset M \bigcup N$, it follows that from property (i) that $[M] \subset [M \bigcup N]$ and $[N] \subset [M \bigcup N]$. But then,

$[M] \bigcup [N] \subset [M \bigcup N]$

which together with (i) implies that $[M \bigcup N] = [M] \bigcup [N]$.

Proof of property 4:

We observe that given any $M \subset R$,

$[M] = [M \bigcup \phi] = [M] \bigcup [\phi]$ by property (iii).

It follows that $[\phi] \subset [M]$. But this is possible for arbitrary M only if $[\phi] = \phi$. (Alternatively, the set with no elements can have no contact points!). QED.

Definition:

A point $x \in R$ is called a limit point of a set $M \subset R$ if every neighbourhood of x contains infinitely many points of M. The limit point may or may not belong to M. For example, if M is the set of rational numbers in the interval $[0,1]$, then every point of $[0,1]$, rational or irrational, is a limit point of M. (It is said that the rationals are dense in R; we will discuss this in detail later).

A point x belonging to a set M is called an isolated point of M if there is a (“sufficiently small”) neighbourhood of x containing no points of M other than x itself.

6.2 Convergence and limits.

A sequence of points $\{x_{n}, n \in N \}$ in a metric space R is said to converge to a point $x \in R$ if every neighbourhood $O_{\epsilon}(x)$ of x contains all points $x_{n}$ starting from a certain index (more exactly, if, given any $\epsilon >0$, there is an integer $N_{\epsilon}$ such that $O_{\epsilon}(x)$ contains all points $x_{n}$ with $n > N_{\epsilon}$). The point x is called the limit of the sequence $\{ x_{n}\}$, and we write $x_{n} \rightarrow x$ (as $n \rightarrow \infty$). Clearly, $\{ x_{n}\}$ converges to x if and only if

$\lim_{n \rightarrow \infty} \rho(x,x_{n}) = 0$.

It is an immediate consequence of the definition of a limit of a sequence that

(1) No sequence can have two distinct limits.

(2) If a sequence $\{ x_{n}\}$ converges to a point x, then so does every subsequence of $\{ x_{n}\}$.

(Details are filled in here: I believe it is helpful to write little proofs of little details…I call it playing with small pearls of math…: )

Proof of (1) : Let, if possible, a sequence $\{ x_{n}\}$ have two distinct limit points, namely, $x_{0}$ and $x_{0}^{'}$. Then, there exists a neigbourhood or open ball $O_{\epsilon/2}(x)$ which contains all points of the given sequence after some $N_{0} \in \mathscr{N}$. So also there exists a neighbourhood (again of radius $\epsilon/2$ about $x_{0}^{'}$ such that it contains all points of the given sequence after some natural number $N_{0}^{'} \in \mathscr{N}$.

But, we know that $\rho(x_{0},x_{0}^{'}) \leq \rho(x_{0},x) + \rho(x_{0}^{'},x) \leq \epsilon/2+\epsilon/2=\epsilon$. Which in turn means that the two “distinct limit points” lie in the same open ball. Hence, they cannot be distinct. Hence, a sequence can have only one limit point, and it is unique. QED.

Proof of (2):

Definition of subsequence: Given a sequence $\{ x_{n}\}$, where $n \in \mathscr{N}$, consider a strictly increasing sequence of positive integers (that is, $n_{1}, n_{2}, n_{3}, \ldots$ are positive integers and $n_{1}) : then the sequence $x_{n_{1}}, x_{n_{2}}, x_{n_{3}}, \ldots$ is called a subsequence of the original sequence $\{ x_{n}\}$.

Now, the claim is that: if a given sequence $\{ x_{n}\}$ converges to a limit point, so does every subsequence of the given sequence: —-

So, it is given that $\rho(x_{n},x) < \epsilon$ when $n > N$ for some natural number $N \in \mathscr{N}$. Since $n_{m} \geq m$ ( here m is some dummy variable, some appropriate positive integer) for all positive integers m, we have $\rho(x, x_{n_{m}})<\epsilon$ whenever $m > N$. This means that $\rho(x_{n_{m}},x)<\epsilon$, which is what we wanted to prove. QED.

THEOREM 2:

A necessary and sufficient condition for a point x to be a contact point of a set M is that there exists a sequence $\{ x_{n}\}$ of points of M converging to x.

PROOF 2:

The condition is necessary since if x is a contact point of M, then every neighbourhood $O_{\frac{1}{n}}(x)$ contains at least one point $x_{n} \in M$, and these points form a sequence $\{ x_{n}\}$ converging to M. The sufficiency is obvious. QED.

THEOREM 2′:

A necessary and sufficient condition for a point x to be a limit point of a set M is that there exists a sequence $\{ x_{n}\}$ of distinct points of M converging to x.

Proof 2′:

Clearly, if x is a limit point of M, then the points $x_{n} \in O_{\frac{1}{n}}(x) \bigcap M$ figuring in the proof of Theorem 2 (just above) can be chosen to be distinct. This proves the necessity and the sufficiency is again obvious. QED.

6.3 Dense subsets, Separable spaces:

Let A and B be two subsets of a metric space R. Then, A is said to be dense in B if $[A] \supset B$. In particular, A is said to be everywhere dense (in R) if $[A] = R$. A set A is said to be nowhere dense if it is dense in no (open) sphere at all. (Note; it is w.r.t. to the metric function defined in the space R.) (Famous example: The rationals are dense in $\Re$).

Example 1:

The set of all rational points is dense in the real line $\Re$.

proof 1:

We want to prove that $[Q] \supset \Re$. Clearly, as $Q \subset R$, the set of rationals are dense in rationals themselves. We only need to show that any irrational number is also a limit point of a sequence of rationals. For example, this is crystal clear: consider the sequence of decimal numbers (finitely many decimals) that we get by extracting square root of 2 by long division. The same applies to any other surd or irrational number. Hence, we can always find a rational sequence of numbers converging to an irrational number. Hence, the rationals are dense in the reals. QED.

Example 2:

The set of all points $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ with rational co-ordinates is dense in each of the spaces $R^{n}$, $R_{0}^{n}$, $R_{1}^{n}$ introduced in the following examples (in earlier blog, and presented here again for handy reference):

2a’) the set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho(x,y) = \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}}$ is a metric space denoted by $R^{n}$ and called n-dimensional Euclidean space (or, simply Euclidean n-space).

proof 2a’:

same proof as example 1. QED.

2b) the set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho_{1}(x,y) = \Sigma_{k=1}^{n}|x_{k}-y_{k}|$ is a metric space denoted by $R_{1}^{n}$.

proof 2b:

same proof as example 1. QED.

2c) The set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}|$ is a metric space denoted by $R_{0}^{n}$.

proof 2c:

once again same proof as example 1 with minor differences. QED.

Example 3:

The set of all points $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{k}, \ldots)$ with only finitely many non-zero co-ordinates, each a rational number, is dense in the space $l_{2}$ introduced in earlier blog (and presented here again for handy reference):

Definition of $l_{2}$ metric space: Let $l_{2}$ be the set of all infinite sequences $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{k}, \ldots)$ of real numbers $x_{1}, x_{2}, \ldots, x_{k}, \ldots$ satisfying the convergence condition $\Sigma_{k=1}^{\infty} x_{k}^{2}< \infty$, where distance between points is defined by $\rho(x,y) = \sqrt{\Sigma_{k=1}^{\infty}(x_{k}-y_{k})^{2}}$.

Again, some detailed steps show that the  essence of this proof is also the fact that the rationals are dense in the reals.

Example 4: So, also, similarly, the set of all polynomials with rational coefficients is dense in both spaces $C_{[a,b]}$ and $C_{[a,b]}^{2}$. Let me reproduce the definitions of these spaces: The set $C_{[a,b]}$ of all continuous functions defined on the closed interval $[a,b]$ with distance $\rho(f,g)= \max_{a \leq t \leq b}|f(t)-g(t)|$ is a metric space. The set of all functions continuous on the interval $[a,b]$ with distance function defined by $\rho(x,y) = (\int_{a}^{b}[x(t)-y(t)]^{2}dt)^{\frac{1}{2}}$ is also a metric space denoted by $C_{[a,b]}^{2}$.

Definition:

A metric space is said to be separable if it has a countable everywhere dense subset.

Example 5:

The spaces $\Re^{1}$, $\Re^{n}$, $\Re_{0}^{n}$, $l_{2}$, $C_{[a,b]}$ and $C_{[a,b]}^{2}$ are all separable since the sets in Examples 1 to 4 above are all countable. (this can be easily verified).

Example 6:

The “discrete space” M described in Example 1 (reproduced here again) contains a countable everywhere dense subset and hence is separable if and only it is itself a countable set, since clearly $[M]=M$ in this case. (Put $\rho(x,y)=1$, when $x=y$ and $\rho(x,y)=0$ if $x \neq y$. Then, this space M is called the discrete metric space).

Example 7:

There is no countable everywhere dense set in the space m of all bounded sequences, introduced in an earlier blog ( To help recall, we present it here again: Consider the set of all bounded infinite sequences of real numbers $x = (x_{1}, x_{2}, \ldots, x_{n}, \ldots)$, and let $\rho(x,y) = \sup_{k}|x_{k}-y_{k}|$. This gives a metric space which we denote by m. The fact that it has the three properties of a metric is almost obvious.)

In fact, consider the set E of all sequences consisting exclusively of zeros and ones. Clearly, E has the power of continuum since there is a one-to-one correspondence between E and the set of all subsets of the set $\mathcal{Z_{+}} = \{1,2, \ldots, n, \ldots \}$ (what is the correspondence here? answer:) Now, the distance between any two points of E equals 1. Suppose we surround each point of E by an open sphere of radius $\frac{1}{2}$, thereby obtaining an uncountably infinite family of pairwise disjoint spheres. Then, if some set M is everywhere dense in m, there must be at least one point of M in each of the spheres. It follows that M cannot be countable and hence, that m cannot be separable.

Closed Sets:

We say that a subset M of a metric space R is closed if it coincides with its own closure, that is, if $[M]=M$. In other words, a set is called closed if it contains all its limit points (we will solve this problem as an exercise).

Example 1:

The empty set $\phi$ and the whole space R are closed sets.

Example 2:

Every closed interval $[a,b]$ on the real line is a closed set.

Example 3:

Every closed sphere in a metric space is a closed set. In particular, the set of all functions f in the space $C_{[a,b]}$ such that $|f(t)|\leq K$ (where K is a constant) is closed.

Example 4:

The set of all functions f in $C_{[a,b]}$ such that $|f(t)| (an open sphere) is not closed. The closure of the set is the closed sphere in the preceding example.

Example 5:

Any set consisting of a finite number of points is closed.

Theorem 3:

The intersection of arbitrary number of closed sets is closed. The union of a finite number of closed sets is closed.

Proof 3:

Given arbitrary sets $F_{\alpha}$ indexed by a parameter $\alpha$, let x be a limit point of the intersection $F = \bigcap_{\alpha}F_{\alpha}$

Then, any neighbourhood $O_{\epsilon}(x)$ contains infinitely many points of F, and hence, infinitely many points of each $F_{\alpha}$. Therefore, x is a limit point of each $F_{\alpha}$ and hence belongs to each $F_{\alpha}$, since the sets $F_{\alpha}$ are all closed. It follows that $x \in F_{\alpha}$ and hence that F itself is closed.

Next, let $F = \bigcup_{k=1}^{\infty}F_{k}$ be the union of a finite number of closed sets $F_{k}$, and suppose x does not belong to F. Then, x does not belong to any of the sets $F_{k}$, and hence, x cannot be a limit point of any of them. But then, for every k, there is a neighbourhood $O_{\epsilon_{k}}(x)$ containing no more than a finite number of points of $F_{k}$. Choosing

$\epsilon = \min (\epsilon_{1}, \epsilon_{2}, \ldots, \epsilon_{n})$

we get a neighbourhoood $O_{\epsilon}(x)$ containing no more than a finite number of points of F, so that x cannot be a limit point of F. This proves that a point $x \in F$ cannot be a limit point of F. Therefore, F is closed. QED.

6.5 Open sets.

A point x is called an interior point of a set M if x has a neighbourhood $O_{\epsilon}(x) \subset M$, that is, a neighbourhood consisting entirely of points of M. A set is said to be open if all points are interior points (with respect to that metric).

Example 1:

Every open interval $(a,b)$ on the real line is an open set. In fact, if $a < x < b$, choose $\epsilon = \min(x-a, b-x)$. Then, clearly $O_{\epsilon}(x) \subset (a,b)$.

Example 2:

Every open sphere $S(a,r)$ in a metric space is an open set. In fact, $x \in S(a,r)$ implies that $\rho(a,x). Hence, choosing $\epsilon = r - \rho(a,x)$ we have $O_{\epsilon}(x)=S(x,\epsilon) \subset S(a,r)$.

Example 3:

Let M be the set of of all functions f in $C_{[a,b]}$ such that $f(t) < g(t)$, where g is a fixed function in $C_{[a,b]}$. Then, M is an open subset of $C_{[a,b]}$.

Theorem 4:

A subset M of a metric space R is open if and only its complement $R-M$ is closed.

Proof of theorem 4:

If M is open, then every point $x \in M$ has a neighbourhood (entirely) contained in M. Therefore, no point $x \in M$ can be a contact point of $R - M$. In other words, if x is a contact point of $R-M$ then $x \in R-M$, that is, $R-M$ is closed.

Conversely, if $R-M$ is closed, then any point $x \in M$ must have a neighbourhood contained in M; since, otherwise every neighbourhood of x would contain points of $R-M$, that is, x would be a contact point of $R-M$, not in $R-M$. Therefore, M is open.

QED.

corollary to theorem 4:

The empty set $\phi$ and the whole space R are open sets.

Proof:

An immediate consequence of Theorem 4 and Example 1 in the previous section.

QED.

THEOREM 5:

The union of an arbitrary number of open sets is open. The intersection of a finite number of open sets is open.

Proof of theorem 5:

this is the dual of theorem 3. The proof is an immediate consequence of theorem 4 and DeMorgan laws. QED.

6.6 Open and closed sets on the real line:

The structure of open and closed sets in a given metric space can be quite complicated. This is true even for open and closed sets in a Euclidean space of two or more dimensions. In the one-dimensional case, however, it is an easy matter to give a complete description of all open sets (and, hence, of all closed sets).

Theorem 6:

Every open set G on the real line is the union of a finite or countable system of pairwise  disjoint open intervals.

Proof of theorem 6:

Let x be an arbitrary point of G. By the definition of an open set, there is at least one open interval containing x and contained in G. Let $I_{x}$ be the union of all such open intervals. Then, as we now show,

$I_{x}$ is itself an open interval.

proof:

In fact, let $a = \inf {I_{x}}$ and $b = \sup {I_{x}}$ where we allow the cases $-\infty=a$ and $+\infty=b$.

Then, obviously, $I_{x} \subset (a,b)$….call this II.

Moreover, let us assume that y is an arbitrary point of $(a,b)$ distinct from x, where, to be explicit, we assume that $a. Then, there is a point $y^{'}\in I_{x}$ such that $a < y^{'}. (HW: think, why!) Hence, G contains an open interval containing the points $y^{'}$ and $x$. But, then this interval also contains y, that is, $y \in I_{x}$. (The case $y>x$ is treated similarly. )

Moreover, $x \in I_{x}$, by hypothesis. It hence follows that $I_{x} \supset (a,b)$, and hence, by (II) that $I_{x}=(a,b)$. Thus, $I_{x}$ is itself an open interval, as asserted, in fact the open interval $(a,b)$.

Next, we want to show that $I_{x}$ is a union of countable, pairwise, disjoint open intervals. Let us proceed as follows:

By its very construction, the interval $(a,b)$ is contained in G and is not a subset of a larger interval contained in G. Moreover, it is clear that two intervals $I_{x}$ and $I_{x^{'}}$ corresponding to distinct points x and $x^{'}$ either coincide or else are disjoint (otherwise, $I_{x}$ and $I_{x^{'}}$ would both be contained in a larger interval $I_{x} \bigcup I_{x^{'}} = I \subset G$. There are no more than countably many such pairwise disjoint intervals $I_{x}$. In fact, choosing an arbitrary rational point in each $I_{x}$ we establish a one-to-one correspondence between the intervals $I_{x}$ and a subset of the rational numbers. Finally, it is obvious that

$G = \bigcup_{x}I_{x}$.

QED.

COROLLARY TO PROOF 6:

Every closed set on the real line can be obtained by deleting a finite or countable system of pairwise disjoint intervals from the line.

EXAMPLE 1:

Every closed interval $[a,b]$ is a closed set (here, a and b are necessarily finite).

EXAMPLE 2:

Every single element set $\{ x_{k}\}$ is closed.

EXAMPLE 3:

The union of a finite number of closed intervals and single element sets is a closed set.

EXAMPLE 4 (The CANTOR Set):

A more interesting example of a closed set on the real line can be constructed as follows:

Delete the open interval $(\frac{1}{3},\frac{2}{3})$ from the closed interval $F_{0}=[0,1]$ and let $F_{1}$ denote the remaining closed set, consisting of two closed intervals. Then, delete the open intervals $(\frac{1}{9}, \frac{2}{9})$ and $(\frac{7}{9}, \frac{8}{9})$ from $F_{1}$ and let $F_{2}$ denote the remaining closed set consisting of four closed intervals. Then, delete the “middle third” from each of these four intervals, getting a new closed set $F_{3}$, and so on. Continuing this process indefinitely, we get a sequence of closed sets $F_{n}$ such that

$F_{0} \supset F_{1} \supset F_{2} \supset \ldots \supset F_{n} \supset \ldots$

(such a sequence is said to be decreasing). The intersection

$F = \bigcap_{n=0}^{\infty}F_{n}$

of all these sets is called the CANTOR SET.

Clearly, F is closed, and is obtained from the unit interval $[0,1]$ by deleting a countable number of open intervals. In fact, at the nth stage of construction, we delete $2^{n-1}$ intervals, each of length $\frac{1}{3^{n}}$.

To describe the structure of the set $F_{n}$ we first note that F contains the points, 0,1, $\frac{1}{3}$, $\frac{2}{3}$, $\frac{1}{9}$, $\frac{2}{9}$, $\frac{7}{9}$, $\frac{8}{9}$, $\ldots$,….call this III.

that is, the end points of the deleted intervals (together with the points 0 and 1)

However, F contains many other points.

In fact, given any $x \in [0,1]$, suppose we write x in ternary notation, representing x as a series :

$x=\frac{a_{1}}{3}+\frac{a_{2}}{3^{2}}+\frac{a_{3}}{3^{3}}+\ldots + \frac{a_{n}}{3^{n}}+\ldots$…call this IV.

where each of the numbers $a_{1}, a_{2}, \ldots$ is 0, 1 or 2 in radix-3. Then, it is easy to see that x belongs to F if and only if x has a representation (IV) such that none of the numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ equals 1 (you need to think about this as HW!!)

Remarkably enough, the set F has the power of the continuum, that is, there are as many points in F as in the whole interval $[0,1]$, despite the fact that the sum of the lengths of the deleted intervals equals

$\frac{1}{3} + \frac{2}{3^{2}} + \frac{2^{2}}{3^{3}} + \ldots =1$

To see this, we associate a new point

$y=\frac{b_{1}}{2} + \frac{b_{2}}{2^{2}}+\ldots + \frac{b_{n}}{2^{n}}+\ldots$

with each point IV, where

$b_{n}=0$, if $a_{n}=0$; $b_{n}=1$; if $a_{n}=2$.

In this way, we set up a one-to-one correspondence between F and the whole interval $[0,1]$. It follows that F has the power of the continuum, as asserted. Let $A_{1}$ be the set of points III. Then,

$F = A_{1} \bigcup A_{2}$ where the set $A_{2}=F-A_{1}$ is uncountable, since $A_{1}$ is countable and F itself is not. The points of $A_{1}$ are often called “points of F of the first kind,” while those of $A_{2}$ are called “points of the second kind.”

QED.

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