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2.5 Basic Concepts |

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2.7 Complete Metric Spaces |

An open sphere/ball $S(x_0, r)$ is the set of points $x\in X$ satisfying the inequality: $$ \rho(x_0, x) < r. $$ where $\rho$ is a metric. The fixed point $x_0$ is the center of the sphere, and $r$ is the radius.

An closed sphere/ball $S[x_0, r]$ is the set of points $x\in X$ satisfying the inequality: $$ \rho(x_0, x) \leq r. $$ where $\rho$ is a metric. The fixed point $x_0$ is the center of the sphere, and $r$ is the radius.

A point $x\in X$ is a contact point for a set $M\subset X$ if every neighborhood of $x$ (epsilon-ball around $x$) contains at least one point of $M$. (For instance, the point 1 is a contact point for $(0,1)$).

The set of all contact points is called the closure of $M$ and denoted by $[M]$. For instance, $[(0,1)] = [0,1]$.

Theorem 1

The closure operator $[\cdot]$ has the following properties:

1) If $M\subset N$ then $[M]\subset[N]$

2) $[[M]] = [M]$

3) $[M\cup N] = [M]\cup[N]$

4) $[\emptyset] = \emptyset$

A sequence $\{x_n\}$ in a metric space converges to a point $x\in X$ if every neighborhood $O_\eps(x)$ contains all points of the sequence after some $N\in\N$. The point $x$ is called the limit of the sequence $\{x_n\}$.

Theorem 2

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

Theorem 2'

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

Let $A$ and $B$ be two subsets of $X$. $A$ is dense in $B$ if $[A]\supset B$. And $A$ is said to be dense everywhere if $[A] = X$.

(Example: $\Q$ is dense everywhere in $\R$.)

Definition

A metric space is said to be

A subset $M$ of a metric space $(X,\rho)$ is

Theorem 3

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

A point $x$ is called an interior point of a set $M$ if $x$ has a neighborhood $O_\eps(x)\subset M$, i.e. it has a neighborhood consisting entirely of points of $M$. A set is said to be

Theorem 4

A subset $M$ of a metric space $(X, \rho)$ is open if and only if its complement $X-M$ is closed.

Corollary

The empty set $\emptyset$ and the whole space $X$ are open sets. (Note: they also satisfy the conditions for being closed sets, so they are both).

Theorem 5

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

Theorem 6

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

Corollary

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

Let $f$ be a mapping from $X$ to $Y$ (metric spaces). The function $f$ is continuous if for all $\eps > 0$ there exists some $\delta > 0$ such that $$ \rho(x_0, x) < \delta \;\Longrightarrow\; \rho'(f(x_0), f(x)) < \eps. $$

Let $X = U(1)\subset\R^2$ where $U(1)$ denotes the unit circle, i.e. the set $\{(x,y)\in\R\,|\, x^2 + y^2 < 1\}$. The distance is the standard Euclidean metric: $$ \rho(x,y) = \sqrt{x^2 + y^2} $$ Now, select two points $x = y = (0,0)$, and set $r_1 = 2$ and $r_2 = 1$. Then $r_1 > r_2$, but $S(x,r_1)\subset S(y, r_2)$. (The sets are restricted to a subspace, and in that subspace they are the same).

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By Theorem 2, for some contact point $y$ for $M$, there exists a sequence $\{x_n\}$ of points in $M$ that converges to $y$. By the definition of convergence, for every epsilon-ball $O_\eps(y)$ there is some $N\in\N$ such that all points in $\{x_n\}$ after $x_N$ lie inside $O_\eps(y)$. There are two exhaustive possibilities,

1) Every $O_\eps(y)$ contains a single point, in which case $y$ is an isolated point.

2) Every $O_\eps(y)$ contains an infinite amount of points, in which case $y$ is a limit point.

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Using the identity from Problem 1 in section 2.5: $$ |\rho(x, z) - \rho(y, u)| \leq \rho(x, y) + \rho(z, u). $$ In our case, we set $x = x_n$, $z = y_n$, $y = x$ and $u = y$: $$ |\rho(x_n, y_n) - \rho(x, y)| \leq \rho(x_n, x) + \rho(y_n, y). $$ Taking the limit: $$ \lim_{n\rightarrow\infty}|\rho(x_n, y_n) - \rho(x, y)| \leq \lim_{n\rightarrow\infty} \Big(\rho(x_n, x) + \rho(y_n, y)\Big) = 0 $$ Which leaves us with: $$ \lim_{n\rightarrow\infty}|\rho(x_n, y_n) - \rho(x, y)| \leq 0 $$ which means $\rho(x_n, y_n)\rightarrow\rho(x, y)$ as $n\rightarrow\infty$.

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Showing implication both ways.

$\Rightarrow$) Assume $f$ is continuous and $x_n\rightarrow x_0$. For any $\eps > 0$, then $\rho'(f(x_n), f(x_0)) < \eps$ since, by continuity of $f$, there exists some $\delta > 0$ such that $\rho(x_n, x_0) < \delta$, which follows by convergence of $\{x_n\}$ (set $\delta = \eps_2 > 0)$. This shows that $f(x_n)\rightarrow f(x_0)$.

$\Leftarrow$) Assume that $x_n\rightarrow x_0$ and that $f(x_n)\rightarrow f(x_0)$ as $n\rightarrow\infty$. For any $\eps > 0$ we can find some $N_1\in\N$ such that $\rho'(f(x_n), f(x_0)) < \eps$ whenever $n\geq N_1$ and some $N_2\in\N$ such that $\rho(x_n, x_0) < \eps$ whenever $n\geq N_2$. Let $\delta = \eps$, then $\rho(x_n, x_0) < \delta$ implies that $\rho'(f(x_n), f(x_0)) < \eps$ for $n\geq \max(N_1, N_2)$ and we can conclude that $f$ is continuous.

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a) The closure of any set $M$ is closed.

Any closed set coincides with its closure, and since by Theorem 1, $[[M]] = [M]$, then $[M]$ is a closed set.

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b) $[M]$ is the smallest closed set containing $M$.

Assume $M$ is a closed set. Then $[M] = M$ which obviously is the smallest set containing $M$.

Now assume that $M$ is an open set. Let $C$ be some arbitrary closed set such that $M\subset C$. Assume $y\in[M]$, then it is a limit point for $M$. Since $M\subset C$ and a closed set contains all its limit points, it follows that $y\in C$, which shows that $[M]\subset C$. Since any closed set $C$ containing $M$ also contains $[M]$, it shows that $[M]$ is the smallest closed set.

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The countable union of closed sets can be an open set, as shown in Problem 7, in section 1.1. $$ \bigcup_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right] = (a, b) $$ And the countable intersection of open sets can be a closed sets, also shown in Problem 7. $$ \bigcup_{n=1}^\infty \left(a - \frac{1}{n}, b + \frac{1}{n}\right) = [a, b] $$

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As noted in the text, any $x\in[0,1]$ can be written in ternary notation: $$ x = \frac{a_1}{3} + \frac{a_2}{3^2} + \ldots + \frac{a_2}{3^n} + \ldots, $$ where $a_i$ can only take values $0, 1, 2$. The $x$ belongs to $F$ if and only if none of the $a_i$ equals 1. If write 1/4 in this form it becomes: $$ \frac{1}{4} = \frac{0}{3} + \frac{2}{3^2} + \frac{0}{3^3} + \frac{2}{3^4} + \frac{0}{3^5} + \frac{2}{3^6} + \ldots, $$ or in decimal form: $$ \frac{1}{4} = 0.0202020202\ldots $$ Since there are no 1s, this shows that $1/4\in F$.

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(a) The points of the first kind; $$ 0, 1, \frac{1}{3}, \frac{2}{3}, \frac{1}{9}, \frac{2}{9}, \frac{7}{9}, \frac{8}{9},\ldots $$ form an everywhere dense subset of $F$.

Define the set of points to be $G$. We want to show that $F\subset [G]$, and we do that by showing that any point in $F$ is a limit point of $G$.

Assume $x\in F$. For any $\eps > 0$ we will consider the $\eps$-ball, which in this case is the interval: $(x - \eps, x + \eps)$. In this interval, we can find some endopints from $G$ satisfying: $$ x - \eps < \frac{3k}{3^n} < \frac{3k+1}{3^n} < x + \eps $$ for some $k,n\in\N$ (a similar argument can be used for any expression of the general formula of the Cantor endpoints). But once we have identified some endpoints that are in the set, we can find an infinite amount of endpoints that are between them, from the Cantor set construction. This means $(x-\eps, x+\eps)$ contains an infinite amount of points from $G$, which means that $x\in F$ is a limit point for $G$.

The closure of a set will contain all its limit points, so since $x\in F$ is a limit point for $G$, then $x\in[G]$. Since $x$ was an arbitrary point, this shows that $F\subset [G]$, which also shows that $G$ is an everywhere dense subset of $F$.

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(b) The numbers of the form $t_1 + t_2$, where $t_1,t_2\in F$, fill the whole interval $[0,2]$.

(a) $x\in A$ implies $\rho(A, x) = 0$, but not conversely.

Assume $x\in A$. Since the distance is nonnegative, $\inf_{a\in A}\rho(a, x)\geq 0$ then $a = x$ so $\inf_{a\in A}\rho(a, x) = \rho(x, x) = 0$.

To show that the converse is not true consider the following counterexample. Let $X = \R$, let $A = (0,1)$ and let $x = 1$. Then $x\not\in A$, but for any $\eps > 0$ we can find some $a\in A$ such that $|a - 1| < \eps$, which means $\inf_{a\in A}\rho(a, x) = 0$.

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(b) $\rho(A, x)$ is a continuous function of $x$ (for fixed $A$).

Defining $f(x) := \rho(x, A)$.

If $V\subset X$ is an open set containing $f(x)$, then $f$ is continuous if there is some open set $U\subset X$ such that $x\in U$ and $f(U)\subset V$.

Assume $f(x)\in V$ for some open set $V$. Then for any $\eps > 0$ there exists an epsilon-ball such that $O_\eps(f(x))\subset V$. Now, our claim is that there exists some open set $U = O_{\frac{\eps}{2}}(x)$ such that $f(U) \subset O_\eps(f(x))\subset V$. Let $y\in A$ and $z\in O_{\frac{\eps}{2}}(x)$, which means that the distance from $z$ to $x$ is smaller than $\eps/2$. By the triangle inequality: $$ \rho(y, z) \leq \rho(x, y) + \rho(x, z) $$ $$ \rho(x, y) \leq \rho(x, z) + \rho(z, y). $$ By the property of the infimum; $$ \rho(z, A) \leq \rho(x, A) + \rho(x, z) < \rho(x, A) + \frac{\eps}{2} $$ $$ \rho(x, A) \leq \rho(x, z) + \rho(z, A) < \rho(z, A) + \frac{\eps}{2} $$ Or directly, $$ \rho(z, A) < \rho(x, A) + \frac{\eps}{2} $$ $$ \rho(x, A) < \rho(z, A) + \frac{\eps}{2} $$ Which, when combined, shows us that: $$ \rho(x, A) - \frac{\eps}{2} < \rho(z, A) < \rho(x, A) + \frac{\eps}{2} $$ which shows that $f(z)\in O_\eps)f(x) \subset V$, and that $f(U)\subset V$ which proves that $f$ is continuous.

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(c) $\rho(A, x) = 0$ if and only if $x$ is a contact point of $A$.

$\Rightarrow$) Assume $\rho(A, x) = 0$. Then for any $\eps > 0$, the point $O_\eps(x)$ will contain points in $A$ and is therefore a contact point of $A$.

$\Leftarrow$) Assume $x\in A$ is a contact point. Assume further that $A$ is closed. Then $x\in A$, since a closed point contains all its contact points. Then by (a), the distance from $A$ to $x$ is 0. Assume now that $A$ is open. Let $x$ be a contact point. By definition, for any $\eps > 0$ the ball $O_\eps(x)$ will contain points from $A$, so $\rho(A, x) < \eps$. Since this is true for any $\eps > 0$, this means that $\rho(A, x) = 0$.

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(d) $[A] = A\cup M$, where $M$ is the set of all points $x$ such that $\rho(A, x) = 0$.

$\subset$) Assume $x\in[A]$. Since $[A]$ is closed it contains all its contact points, so $\rho(A, x) = 0$, which means $x\in M \subset A\cup M$. Hence $[A]\subset A\cup M$.

$\supset$) Assume $x\in A$. Since $A\subset[A]$ then $x\in [A]$. Assume now that $x\in M$ and $x\not\in A$. Then by definition of $M$, $\rho(A, x) = 0$ which means $x$ is a contact point for $A$ by (c), which also means $x\in[A]$. It follows that $A\cup M\subset[A]$.

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Let $x\in A\cap B$. Then by 9(a), $\rho(x, A) = 0$ and $\rho(x, B) = \rho(B, x) = 0$. From the triangle inequality, $\rho(A, B) \leq \rho(A, x) + \rho(x, B) = 0 + 0 = 0$ which shows that $\rho(A, B) = 0$.

Showing that the converse is not true with a counterexample. Let $X = \R$, let $A = (0,1)$ and let $B = [1, 2)$. Then as shown above, $\rho(A, 1) = 0$ and since $1\in B$, $\rho(A, B) = 0$.

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