1 Prelimaries

This section is a dump of various preliminary results for reference. Please see the next section for the main content of the course.

Topology

First recall some topological definitions:

  1. The norm \(|x|\) of \(x\) equals \((\sum x_j^2)^{1/2}\).
  2. The distance between two points is is \(|x-y|\).
  3. The distance between two sets is \(d(E, F)=\inf_{x\in E, y\in F}|x-y|\).
  4. A set \(E\) is bounded if it is contained in some ball of finite radius.
  5. A bounded set is compact if it is also closed.
  6. A point \(x\) is a limit point of the set \(E\) if \(\forall r>0, B_r(x)\cap E\neq \emptyset\).
  7. \(x\in E\) is isolated if \(\exists r>0: B_r(x)\cap E=\{x\}\).
  8. \(x\in E\) is an interior point if \(\exists r>0: B_r(x)\subset E\); the interior points of \(E\) is denoted \(\mathrm{Int}(E)\).
  9. The boundary \(\partial E = \bar E - \mathrm{Int}(E)\).
  10. A closed set \(E\) is perfect if it does not have any isolated points.

Proposition 1.1 A closed subset \(A\subset K\) where \(K\) is compact, is compact.

Proof: \(V=K-A\) is open (if \(K\) is a subset, then in the subsace topology), then for any \(\{U_j\}\) covering \(A\), \(\{U_j\}\cup \{V\}\) will cover \(K\). Choose a finite subcover and remove \(V\) to obtain a finite subcover for \(A\).

Proposition 1.2 A compact subset \(K\) of a Hausdorff space \(\Omega\) is closed.

Proof We show that \(\Omega-K\) is open: for each \(y\in \Omega-K\), use Hausdorff to construct a open covering \(\{(U_x, V_x)\}_{x\in K}\) separating \(x, y\) such that \(x\in U_x, y\in V_y\). Choose a finite subset of \((U_x)_{x\in J\subsetneq K}\) covering \(K\), then \(y\in \bigcap_{x\in J} V_x \subset \Omega-K\), so \(\Omega-K\) is open.

Proposition 1.3 (closed+bounded=compact) For \(\mathbb R^d\), a set \(A\) is compact if and only if it is closed and bounded.

Proof Given compactness, choose a cover by unit-cube coverings, then \(A\subset [-R, R]^n\), which is bounded. Closure follows from proposition 1.2. Conversely, given \(A\subset [-R, R]^n\) and closed, it is a closed subset of a compact set, so it is compact by proposition 1.1

Lemma 1.1 (finite distance between compact and closed sets) If \(F\) is closed and \(K\) compact, and \(F\cap K = \emptyset\), then \(d(F, K)>0\).

Proof By \(F^c\) open for each \(x\in K\) choose \(B(2\delta_x, x)\subset F^c\). Choose an finite open subcover \(\{B_{\delta_x, x}\}_{x\in [n]}\) of \(K\), then \(d(F, K) > \min_{x\in [n]} \delta_x\).

Proposition 1.4 (compactness and product) \(\Omega, Y\) compact implies \(\Omega\times Y\) compact.

Lemma 1.2 (continuity of distance) Given a metric space \((\Omega, d)\) and a subset \(Y\subset \Omega\), the function \(\varphi:\Omega\to \mathbb R\) defined below is continuous: \[ \varphi(x) = \inf_{y\in Y} d(x, y) \]

Proof The preimage operation commutes with arbitrary union and intersections, so it suffices to demonstrate this for open intervals \(\{(0, a)\}\cup \{(b, \infty)\}\), to show that \[ \varphi^{-1}((0, a)) = \{ x\in \Omega: \forall \epsilon>0, \exists y\in Y: d(x, y) < a+\epsilon \} \] is open, similarly for the other subbasis component. Standard argument is omitted.

Lemma 1.3 (Lebesgue number lemma) Given a compact metric space \((\Omega, d)\) and an open cover \((U_j)\) of \(\Omega\), there \(\exists\delta>0\) such that any subset of diameter \(<\delta\) is entirely contained in a single open set.

Proof Select a finite subcover \(U_1\cup \cdots \cup U_n\) of \(\Omega\). The function \[ f(x) = \min_j d(x, \Omega-U_j) \] is continuous on compact interval, thus achieves its min \(\delta\).

Definition 1.1 (connected sets) A space \(\Omega\) is connected if any of the following equivalent conditions are true:

  • Every binary partition \(U\sqcup V=\Omega\) of \(\Omega\) is trivial: \(U\) or \(V=\emptyset\) and the other \(\Omega\).
  • The only subsets which are both open and closed in \(\Omega\) is \(\emptyset, \Omega\).

Real analysis

Definition 1.2 (Riemann integral) Fix \([a, b]\) and a partition \(a=x_0\leq \cdots \leq x_n =b\), define \(\Delta x_j = x_j - x_{j-1}\) and \[\begin{align} M_j &= \sup f([x_{j-1} \leq x\leq x_j]), \quad m_j = \inf f([x_{j-1} \leq x\leq x_j]) \\ U(P, f) &= \sum_{j=1}^n M_j \Delta x_j, \quad L(P, f) = \sum_{j=1}^n m_j \Delta x_j \end{align}\] The upper and lower integrals are defined as \[\begin{align} \bar{\int}^b_a f\, d\alpha = \inf U(P, f, \alpha), \quad \underline{\int}^b_a f\, d\alpha = \sup L(P, f, \alpha) \end{align}\]

Semi-continuity

Definition 1.3 (limit inferior) Given a countable sequence of reals \((a_j)\), the limit inferior is \[ \liminf_{n\to \infty} a_n = \lim_{n\to \infty} g_n, \quad g_n = \inf_{k\geq n} a_k \] The counterpart \(\limsup\) is defined similarly.

Definition 1.4 (semi-continuity) A \(\mathbb R^n\)-valued function \(f\) is lower-semicontinuous if \[ f^{-1}((t, \infty)) \text{ is open }\forall t \] It is upper-semicontinuous when \(f^{-1}((-\infty, t))\) is open for all \(t\). Here \((t, \infty)\) denotes the set of all vectors which larger than \(t\) pointwise.

Proposition 1.5 (ε-δ definition of semi-continuity) Given a metric space \(X\), \(f:X\to \mathbb R^n\) is lower semicontinuous at \(x_0\in X\) if for every vector \(\epsilon>0\) there exists a real scalar \(\delta>0\) such that \[ f(B_\delta(x_0)) \geq f(x_0) - \epsilon \] Similarly, it is upper-semicontinuous when \(f(B_\delta(x_0)) \leq f(x_0) + \epsilon\).

Proof: We do the proof for lower-semicontinuity.

  • Assuming definition 1.4, \(f^{-1}((t-\epsilon, \infty))\) open for every \(\epsilon\) implies that at \(x_0\in f^{-1}(t-\epsilon, \infty)\) has an open neighborhood.
  • Conversely, assuming the given condition holds at every \(x_0\in X\), the preimage of every open set \(f^{-1}((t, \infty))\) contains an open neighborhood for each point thus is open.

Proposition 1.6 (limit definition of semi-continuity) \(f:X\to \mathbb R^n\) is lower semicontinuous at \(x_0\) if \[ \liminf_{x\to x_0} f(x) \geq f(x_0) \] Similarly, it is upper-continuous at \(x_0\) if \(\limsup_{x\to x_0} f(x) \leq f(x_0)\).

Proof: \(\liminf_{x\to x_0} f(x) \geq f(x_0)\) implies that \(\forall \epsilon>0\in \mathbb R^n\), there exists \(\delta > 0 \in \mathbb R\) such that \(f(B_\delta(x_0))\geq f(x_0) - \epsilon\). Similarly, \(\limsup_{x\to x_0} f(x) \leq f(x_0)\) if for every \(\epsilon>0\in \mathbb R^n\), there exists \(\delta>0\in \mathbb R\) such that \(f(B_\delta(x_0)) \leq f(x_0) + \epsilon\).

Example 1.1 Of the following, \(f_1, f_4\) is lower but not upper semi-continuous, \(f_2, f_3\) is upper but not lower semi-continuous. is \[ f_1(x) = \begin{cases} 0 & x\leq 0 \\ 1 & x > 0 \end{cases}, \quad f_2(x) = \begin{cases} 0 & x < 0 \\ 1 & x \geq 0 \end{cases}, \quad f_3(x) = \begin{cases} 0 & x \leq 0 \\ -1 & x > 0 \end{cases}, \quad f_4(x) = \begin{cases} 0 & x < 0 \\ -1 & x \geq 0 \end{cases} \] (Lower, upper) semi-continuous functions can only jump (down, up) in limit on the side with open iterval.

Cauchy sequences

Definition 1.5 (Cauchy sequence) A sequence \(f_j\) in a metric space \((V, d)\) is a Cauchy sequence if \(\forall \epsilon>0\), there exists \(n\in \mathbb Z_+\) such that \(d(f_{\forall j\geq N}, f_{\forall k\geq n})<\epsilon\).

Proposition 1.7 (convergent subsequence) Given a Cauchy sequence \((f_j)\), if some subsequence \(f_{n_j}\to f\) converges, then the whole sequence \(f_j\to f\) converges.

Proof (sketch): Use the triangle inequality and convergence condition.