[Calculus] 4. Metric Space

Proof By definition: \[\begin{align*} d(x,y) & \le d(x,z) + d(y,z)\\ d(z,y) & \le d(x,y) + d(x,z) \end{align*}\\ \Darr \\ \begin{align*} d(x,y) - d(y,z) & \le d(x,z) \\ -d(x,z) & \le d(x,y) - d(y,z) \end{align*}\\ \]

That is expected. □

Proof Proof 1.

For any \(x\in X\), we have \(d(x,x_0) < r\). Let \(r' = \frac{r - d(x,x_0)}{2} > 0\), consider the open ball \(B_{r'}(x)\), \(\forall a \in B_{r'}(x)[d(a,x) < r']\).

By triangle inequality (definition of metric)

\[d(a,x_0) \le d(x,x_0) + d(a,x) < d(x,x_0) + r' = \frac{r + d(x,x_0)}{2} < r\]

Therefore, \(B_{r'}(x) \sube B_r(x_0)\).

Proof 2.

Let \(A := X\backslash\overline{B_r(x_0)} = \{x\in X | d(x,x_0) > r\}\).

For any \(x\in A\), we have \(d(x,x_0) > r\). Let \(r' = \frac{d(x,x_0) - r}{2} > 0\), consider the open ball \(B_{r'}(x)\), \(\forall a \in B_{r'}(x)[-d(a,x) > -r']\).

By triangle inequality (Proposition 4.2)

\[d(a,x_0) \ge |d(x,x_0) - d(a,x)| > |d(x,x_0) - r'| = \left|\frac{d(x,x_0) + r}{2}\right| > r\]

since \(d(x,x_0) > r \Rarr r + d(x,x_0) > 2r\).

Therefore, \(B_{r'}(x) \sube A\). □

Proof Proof 1. Were this false, then we can find an \(x_0 \in \varnothing\)… however, there is not any elements in \(\varnothing\). Thus \(\varnothing\) is open logically.

Proof 2. Too obvious. Beginning with \(\forall x_0 \in O_1\cap O_2\), then it comes to \(x_0 \in O_1\) or \(x_0 \in O_2\), and the rest part is natural.

Proof 3. Still obvious. □

Proof (\(\Rarr\))

For any open set \(U\sube X\), there exists an \(epsilon > 0\), then \(B_\epsilon(L) \sube U\). For this \(\epsilon\), by definition of limit in under concept of metric, there exists an \(N\in\N\), such that \([n\ge N \Rarr d(a_n,L) < \epsilon]\), i.e., \(a_n\in B_\epsilon(L) \sube U\).

(\(\Larr\))

For any real number \(\epsilon > 0\), similarly choose an open set \(U := \{u\in X | d(u,L) < \epsilon\}\). Then there exists \(N\in\N\), such that \([n\ge N \Rarr a_n \in U]\), i.e., \(d(a_n, L) < \epsilon\). □

Proof Choose \(M > 0\), let \(Q_0 := [-M,M]\times[-M,M]\), such that \(a_n\in Q_0\) for all \(n\in\N\). This is possible since \(a_n\) is bounded.

Divide \(Q_0\) into 4 squares with equal size. There must be infinitely many terms inside one square (Were this false, the 4 squares both have finitely many terms, thus the total terms are finite. This is a contradiction).

Choose that square, say \(Q_1\), such that \(\left|\{n|a_n\in Q_1\}\right| = \infty\). Select \(n_1\in\N\) such that \(a_{n_1}\in Q_1\).

Repeating the above operation, we obtain a kind of nested intervals, \(Q_1, Q_2,\cdots\).

We claim that there exists an \(a\in\R\), such that \(\bigcap_n Q_n = \{a\}\).

Now we are facing the squares, which are just 2-dimension of intervals. Apply the theorem of nested intervals into two dimension respectively, we can obtain that \(\bigcap_n Q_n \ne \varnothing\).

And there is only exactly one element is this intersection. Otherwise, there must a case that \(a,a'\) belong to two squares respectively, and no matter which square is selected, it will not get to the other one.

At last, it can be seen that \(\lim_{k\to\infty}a_{n_k} = a\). Given a radius, there must be a square inside that circle after a big enough \(n\). □

Proof (\(\Rarr\))

For producing contradiction, assume that there is an sequence \(a_n(n\in\N)\) in \(F\), such that \(\lim_{n\to\infty} a_n = a\) but \(a\notin F\). i.e., \(a\in X\backslash F\).

\(F\) is closed, then \(X\backslash F\) is open. There exists a radius \(r > 0\), such that \(B_r(a) \sube F\).

Then for all \(n\in \N\),

\[ a_n\in F \Rarr a_n \notin B_r(a) \Lrarr d(a_n, a) > r\]

This contradicts with \(lim_{n\to\infty} a_n = a\).

(\(\Larr\))

For producing contradiction, suppose that \(F\) is not closed, and hence \(X\backslash F\) is not open. Thus

\[\exists a \in X\backslash F \forall r > 0 \exists b \in F[b\in B_r(a)]\]

Assume that \(L\in X\backslash F\) is that point mentioned above as \(a\). Let \(r_n = \frac{1}{n}(n\in\N)\). Then there exists an element in \(F\), say \(a_1\), such that \(a_1 \in B_{r_1}(L)\), and other element, say \(a_2\), such that \(a_2 \in B_{r_2}(L)\).

We collect them all, then obtain a sequence \(a_n(n\in\N)\) such that \(a_n\in B_{r_n}(L)\) for all \(n\in \N\).

For any \(\epsilon > 0\), there exists an \(N\in\N\), such that \(r_N = \frac{1}{N} < \epsilon\), by Archimedean property. And for \(n\ge N\), we have \(d(a_n, L) < r_n < r_N < \epsilon\), i.e., \(\lim_{n\to\infty} a_n = L\).

However, this leads a contradiction since all the element of \(a_n\) is in \(F\), while its limit is not. □

Proof (\(\Rarr\))

\(S\) is compact, then \(S\sube \bigcup_{j=1}^n O_{\alpha_j}\), i.e., \(\bigcup_{j=1}^n O_{\alpha_j}\) is actually an bound of \(S\). Then \(S\) is bounded.

For contradiction, assume that \(S\) is not closed.

Since \(S\) is bounded, then we can find a cube: \(\overbrace{[-M,M] \times \cdots \times [-M,M]}^{m}\), denoted by \(O_0\), such that \(S\sube O_0\).

Divide \(O_0\) into 4 equal-sized cubes. Select that one, say \(C_1\), whose intersection with \(S\) is not closed. (This case must happen, otherwise all cubes' intersection with \(S\) is closed, i.e., \(O_0\cap S = S\) is closed, contradicting with assumption.) And let \(O_1 = O_0 \backslash C1\).

Repeating the operation, we obtain: \(C_1,C_2,\cdots\), and \(O_1,O_2,\cdots\).

Select an sequence \(a_1,a_2\cdots\), where \(a_n\in C_n\) for all \(n\in\N\). By the theorem of nested intervals, \(\lim_{n\to\infty}a_n = a\) for some \(a\in\R^m\).

And we claim that \(O_1,O_2,\cdots\) is an open cover of \(S\). Were this false, \(\exists s\in S\forall n\in\N[s\notin O_n]\), i.e., \(\forall n\in\N[s\in C_n]\). Therefore, \(s = \lim_{n\to\infty} a_n = a\).

By Proposition 4.9, \(S\) is closed. This is a contradicion.

Thus, \(S\) is closed.

(\(\Larr\))

Suppose that \(S\) is bounded and closed, and \(\exists \text{open cover } O_\alpha(\alpha\in A)\) which admits no finite subcover.

Choose a cube \(Q_0\) containing \(S\).

Divide \(Q_0\) into 4 euqal-sized cubes, and select one of them, say \(Q_1\), such that \(Q_1\cap S\) can not be covered by finitely many \(O_\alpha\). (This kind of cube must exist, otherwise all 4 cubes can be coverd by finitely many open sets, also the whole cube, i.e., \(S\) is compact. This contradics with assumption.)

Repeat above operation. Like proof of Bolzano-Weierstrass theorem, we obtain \(Q_1,\cdots, Q_n\), and \(\bigcap_n Q_n = {a}\) for some \(a\in\R^m\).

Choose a seires of points \(s_1,s_2,\cdots\), where \(s_n\in Q_n\cap S\) for all \(n\in\N\). Then \(\lim_{n\to\infty}s_n = a\).

By Proposition 4.9, \(S\) is closed, \(s_n\in S(n\in\N)\), then \(a\in S\).

There exists an \(\alpha\in A\), such that \(a\in O_\alpha\). And we can find an open ball \(B\), such that \(a\in B\sube O_\alpha\).

In above infinitely dividing operations, \(\exists n\in\N[Q_n \sube B\sube O_\alpha]\), i.e., there must be an cube small enough contained inside the open ball.

However, all \(Q_n\) we choose can not be coverd by finitely many open sets, we now have \(Q_n\) is cover by an open set. This is a contradiction, therefore, \(S\) must be compact. □

Proof For every \(p\in K\), there exists \(\alpha(p) \in A\) such that \(p\in O_{\alpha(p)}\). \(O_{\alpha(p)}\) is open, and hence \(\exists r(p) > 0 [B_{r(p)}(p) \sube O_{\alpha(p)}]\).

Then \(B_{\frac{r(p)}{2}}(p)(p\in K)\) form an open cover of \(K\). Thus

\[\exists p_1,\cdots, p_k \in K[ K\sube \bigcup_{j=1}^k B_{\frac{r(p_j)}{2}}(P_j)]\]

Let \(\delta = \min\{\frac{r(p_1)}{2}, \cdots, \frac{r(p_k)}{2}\} > 0\).

For any \(x\in K\), there exists \(p_j(1 \le j \le k)\) such that \(x\in B_{\frac{r(p_j)}{2}}(p_j)\).

For all \(x'\in B_\delta(x))\), we have

\[\begin{align*} d(x', p_j) &\le d(x',x) + d(x, p_j) \\ &< \delta + \frac{r(p_j)}{2} \\ &\le \frac{r(p_j)}{2} + \frac{r(p_j)}{2} = r(p_j) \end{align*}\]

Thus, \(x'\in B_{r(p_j)}(p_j)\sube O_{\alpha(p_j)}\), i.e., \(B_\delta(x) \sube O_{\alpha(p)}\). □

Proof For convenience, let \(A := \{x \in S| A(x)\}, I := \{x \in S | I(x)\}, L := \{x \in S | L(x)\}\).

Proof 1.

Suppose \(L\cap I \ne \varnothing\), i.e., \(\exists x\in S[x\in L \land x\in I]\).

\(x\in L \Rarr [B_\epsilon(x)\cap S\backslash\{x\} \ne \varnothing]\), i.e., \(\exists x'\in S[x'\in B_\epsilon(x) \land x' \ne x]\), which contradicts with \(x\in I \Rarr \exists \epsilon > 0 [B_\epsilon(x) \cap S = \varnothing]\).

Therefore, \(L\cap I = \varnothing\).

Proof 2.

(\(\Rarr\))

For all \(a\in A\), there exists a sequence \(a_n \in S\), such that \(\lim_{n\to\infty} a_n = a\), i.e., \(\forall \epsilon > 0 \exists N\in\N [n \ge N \Rarr d(a_n,a) < \epsilon]\).

Case I: There are infinitely many terms \(a_j\) which do not euqal to \(a\).

There stiall infinite terms \(a_j\) such that \(a_j \ne a\) as \(n \ge N\). And hence we have \(B_\epsilon(a)\cap S\backslash\{a\} = \{a_j| j \ge N \land a_j \ne a}\).

By definition, \(a\) is a limit point, i.e., \(a\in L\).

Case II: There are only finitely many terms which do not equal to \(a\). And assume \(a_M(M\in\N)\) is the last one, i.e., \(\forall n\ge M[a_n = a]\).

Let \(N' = \max\{N, M\}\). Then \(n \ge N' \Rarr d(a_n, a) < \epsilon \land a_n = a\). In this case, \(B_\epsilon(a)\cap S = \{a_n| n \ge M\} = \{a\}\).

By definition, \(a\) is a isolation point, i.e., \(a\in I\).

(\(\Larr\))

If \(a\in I\), just select this sequence \(a,a,a,\cdots\) whose elements are both \(a\). It is clear that it converges to \(a\).

If \(a\in L\), for a given \(\epsilon > 0\), there are points inside \(B_\epsilon(a)\). Collect them all. If they are finite, repeat them in sequence. Then we have a sequence, which is also convergent to \(a\) apparently.

Proof 3.

For contradiction, assume that \(L = \varnothing\).

For all \(x\in S(\sube K)\), there always exists an \(\epsilon > 0\), such that \(B_\epsilon(x) \cap S\backslash\{x\} = \varnothing\). i.e., there are no more else elements inside the ball except center itself.

Collect all that balls, write them as \(O_{\alpha(x)}\) where \(\alpha(x)\) belongs to some index set determined by \(x\in S(\sube K)\).

It is obvious that \(O_{\alpha(x)}\) is an open cover of \(S\). However, we can not cover \(S\) with finite sets, since there are infinitely many elements in \(S\). This contradicts with the fact that \(K\) is compact. □