Analysis, Part I: Metric Spaces

August 19th, 2017

The general idea is to begin by defining the notion of convergence for arbitrary spaces.

Definition: A metric space is a set $X$ equipped with a function $d:X\times X \rightarrow [0, \infty]$ such that:

We now define convergence:

Definition: A sequence $x_n$ converges to $x$ with respect to a metric $d$ if $\underset{n \to \infty}{\lim} d(x_n, x)$ exists and is equal to $0$.

It is not hard to check that a sequence converges in the Euclidean, taxi-cab, or sup norm metric iff each of its components converges.

Lemma

Limits are unique in a metric space.

Since a sequence gets arbitrarily close to its limit, we can prove that if there are two limits, they are also arbitrarily close – and indeed so they must be equal.

Point-Set Topology

Definition: A neighborhood of $x_0$ in $X$ is a set $B(x_0,r) = \{ y\in X \mid d(x_0, y) < r \}$, for some $r>0$. This is sometimes called an open ball.

For later on, we define the useful notion of a limit point.

Definition: A limit point $x$ of $E$ is a point such that every neighborhood of $x$ intersects $E$ in some point other than $x$ itself.

Definition: $x_0 \in \text{int}(E)$, the interior of $E$, if some neighborhood of $x_0$ is contained in $E$.

Definition: $x_0 \in \text{ext}(E)$ if there is some neighborhood of $x_0$ which does not intersect $E$.

Definition: A boundary point of $E$ is neither in its interior or exterior. The set of all boundary points of $E$ is denoted $\partial E$.

It is clear that around a boundary point, every neighborhood intersects both $E$ and the complement of $E$, which is what we would expect.

Definition: The closure $\bar{E}$ of $E$ is the set of points $x\in X$ so that every neighborhood of $x$ has nonempty intersection with $E$; indeed this definition immediately suggests the following proposition.

Proposition

The following are equivalent for $x_0\in X$ in a metric space $X$.

Now, we can define the fundamental concepts of topology: open and closed sets.

Definition: $E$ is closed if it contains its boundary.

Definition: $E$ is open if it does not intersect its boundary. Equivalently, if $E=\text{int}(E)$.

Proposition

All these facts are proven in any topology book, or in my topology notes.

Relative Topology

Note that open sets are determined by both the metric itself and by the choice of ambient space. For example, an open interval of the $x$-axis is open when considered as a part of $\mathbb{R}$, but considered in the plane it is not open.

Definition: Given a metric space $X$ and $Y \subseteq X$, we call $E$ relatively open with respect to $Y$ if it is open with respect to the metric space induced on $Y$ by the metric of $X$.

Proposition

This is exactly what we do when constructing the subspace topology, as discussed in the aforementioned notes.

Cauchy Sequences

Lemma

If a sequence converges, every subsequence converges as well.

If we pick a large enough $N$, then for $n \ge N$, every element of any subsequence will get arbitrarily close to a limit.

Proposition

Suppose we have a sequence $\{x_n\}$ following are equivalent:

The backwards direction is obvious. For the forward direction, we start by constructing a convergent subsequence. For each $N>0$, and each natural number $k$, there exists an integer $M > k$ so that:

$$ d(a_M, L) < \frac{1}{N} $$

So $M(k, N)$ exists for each choice of $k, N$. We define the sequence:

$$ m_0 = 0 \\ m_1 = M(m_0, 1) \\ m_2 = M(m_1, 2) \\ \dots $$

This is a subsequence of $\{x_n\}$ by definition. And for each $\epsilon < \frac{1}{N}$, we have constructed $x_{m_N}$ which lies within distance $\epsilon$ of $L$. Thus, we have found a subsequence which converges to $L$.

We now move on to an important class of sequences.

Definition: $\{x_n\}$ is called a Cauchy sequence if for each $\epsilon > 0$, there exists an integer $N$ so that if $j, k \geq N$, then $d(x_j, x_k) < \epsilon$.

Lemma

A convergent sequence is Cauchy.

For each epsilon, we pick $N_\epsilon$ so that for all $n\geq N_\epsilon$, $d(x_n, L) < \epsilon$. But then by the triangle inequality, for $j,k \geq N_{\epsilon/2}$, we have:

$$ d(x_j, x_k) \leq d(x_j, L) + d(x_k, L) \leq \epsilon $$

Proposition

Every subsequence of a Cauchy sequence is Cauchy.

This proof is very similar to the proof that every subsequence of a convergent sequence converges.

We remark here that not every Cauchy sequence converges, so the converse of the above lemma is not in general true. For example, take the successive truncated decimal approximations of $pi$; this sequence is Cauchy, but it does not converge in $\mathbb{Q}$. We naturally ask, in what kind of a space is a Cauchy sequence always convergent?

Lemma

Let $\{x_n\}$ be a Cauchy sequence in a metric space $X$. Suppose some subequence converges to $L\in X$. Then $\{x_n\}$ converges to $L$.

We noted that if a sequence converges, every subsequence converges. Here, we have something like a converse. If a Cauchy sequence has a convergent subsequence, then the sequence converges as well.

We now explores the conditions under which Cauchy sequences and convergent sequences are equivalent.

Definition: A metric space $X$ is complete if every Cauchy sequence converges in $X$

Propoosition

Let $X$ be a metric space, and let $Y$ be a subspace with the induced metric. Then, if $Y$ is complete, then $Y$ is closed in $X$. Conversely, if $X$ is complete, and $Y$ is closed, then $Y$ is complete.

Closed subsets of complete spaces are complete; complete subsets of arbitrary spaces are closed. In a sense, then, complete metric spaces are intrinsically closed, regardless of the ambient context; and if we take closed subsets of complete spaces, this property is preserved.

To prove the first proposition, suppose $Y$ is complete. Take any convergent sequence in $X$. Since convergent sequences are Cauchy, its limit must be in $Y$. Therefore, $Y$ contains all its limit points and it is closed.

To prove the second proposition, suppose that $Y$ is closed, and $X$ is complete. Take a Cauchy sequence in $Y$. Since $X$ is complete, the sequence converges in $X$. Therefore, the limit of this Cauchy sequence lies in $\bar{Y}$; and since $Y$ is closed, indeed the limit lies in $Y$. Therefore, the Cauchy sequence in $Y$ converges in $Y$.

Note that proving both statements tells us that in a complete metric space, closed sets are complete and vice versa. In general, however, completeness is strictly stronger than the closed condition.

Compact Metric Spaces

Definition: A metric space is compact iff every sequence has a convergent subsequence.

Compactness, unlike some of the other properties mentioned, does not depend on the ambient context. We’ll see why in a bit. First,, we explain some of the conditions associated with compactness.

Definition: Let $Y \subset X$. $Y$ is bounded if there exists an open ball around some point $x\in X$ which entirely contains $Y$.

Proposition

If $X$ is compact, then it is complete and bounded.

Let $\{x_n\}$ be a Cauchy sequence in a compact space. Then, there exists a convergent subsequence. By an earlier lemma, this means that $\{x_n\}$ converges as well. Therefore, $X$ is complete.

To prove the boundedness condition, we use the triangle equality. Pick an arbitrary $x\in X$. If $X$ is not bounded, then we can construct a sequence $\{x_n\}$ so that $d(x_n, x) > n$. Fix another integer $N$. Then, for $m> N$, we have:

$$ d(x, x_m) \leq d(x, x_N) + d(x_m, x_N) \\ d(x, x_m) - d(x, x_N) \leq d(x_m, x_N) $$

Since we fixed $N$, we can pick $m$ large enough that the quantity on the left hand side blows up. It is not hard to see that this sequence has no convergent subsequence, since no subsequence is Cauchy; we fail to meet the condition that $d(x_m, x_N)$ goes to $0$.

Corollary

If $X$ is a metric space, and $Y$ is compact in $X$, then $Y$ is closed and bounded.

Completeness implies closedness, and so we are done.

Theorem

In Euclidean space, a set $E$ is compact iff it is closed and bounded.

So the converse of the above theorem holds true not in general, but in Euclidean spaces. This is the famous Heine-Borel theorem from real analysis. The proof of the converse goes something like this: in a closed and bounded space, every sequence has a monotone subsequence, and such subsequences must converge.

We now have an alternative definition of compactness, which may be considered the “true” definition since it differs in some kinds of topological spaces from the one we constructed earlier. You shouldn’t feel bad if this definition is puzzling; it took a lot of work to find the right condition to discuss these notions.

Theorem

Let $Y \subseteq X$ be a compact subspace of a metric space. Let $V_\alpha$, $\alpha \in A$ be an open cover for $Y$, i.e. a collection of (possibly finite) open sets whose union contains $Y$. Then tehre is a finite subset $F \subseteq A$ so that:

$$ Y \subseteq \bigcup\limits_{\alpha \in F} V_\alpha $$

More succinctly: every open cover of $Y$ has a finite subcover.

Proof

We assume that there exists no finite subcover. Then, we pick $y\in Y$ arbitrarily. We know that some neighborhood of $y$ is contained in $V_\alpha$ for some $\alpha$. We define:

$$ r(y) = \text{sup} \{ r \mid B(y, r) \subset V_\alpha \text{ for some } \alpha \} $$

So $r(y)$ is the size of the largest ball around $y$ which is contained in a single element of the open cover. We define:

$$ r_0 = \text{inf} \{ r(y) \mid y\in Y \} $$

So $r_0$ is the smallest radius for which a ball of size $r_0$ around every $y\in Y$ is contained in a single element of the open cover.

There are two cases here: $r_0 = 0$, or $r_0 > 0$.

Case 1:

Let $r_0=0$. For each positive integer $n$, we can pick (by the axiom of choice) some element $y_n$ so that $r(y_n) < \frac{1}{n}$. The limit of the sequence $\{r(y_n)\}$ must be zero.

Since $Y$ is compact, we have a convergent subsequence $y_{n_k}$, which converges to some $L\in Y$. If we pick small $\epsilon > 0$ we can have $B(L, \epsilon) \subset V_\alpha$ (since $V_\alpha$ is open).

Furthermore, there is some integer $N$ so that for all $n_k > N$, $y_{n_k} \in B(L, \frac{\epsilon}{2})$.

Putting these facts together, we pick an arbitrary $y\in B(y_{n_k}, \frac{\epsilon}{2})$; by the triangle inequality, we have for $n_k \geq N$:

$$ d(y, L) \leq d(y, y_{n_k}) + d(y_{n_k}, L) \\ d(y, L) \leq \epsilon $$

So indeed this means that $B(y_{n_k}, \frac{\epsilon}{2}) \subseteq B(L, \epsilon) \subseteq V_\alpha$. But this means that $r(y_{n_k}) \geq \frac{\epsilon}{2}$ for each $n_k \geq N$. This contradicts our earlier assumption that the limit of the sequence $\{r(y_n)\}$ must be zero.

Case 2:

In this case $r_0 > 0$. So in particular, $r(y) > r_0/2$ for each $y\in Y$. Equivalently, for each $y\in Y$, there is some $\alpha$ so that $B(y, r_0/2) \subseteq V_\alpha$.

We pick $y_1 \in Y$ and construct an open ball around $y_1$ of radius $r_0/2$. This ball cannot cover $Y$, since that would mean we have a finite subcover $V_\alpha$.

Since the above set does not contain $Y$, we can pick $y_2$ from outside the set (i.e. farther than distance $r_0/2$ away from $y_1$) and add to our construction an open ball around $y_2$ of radius $r_0/2$. Our construction still cannot cover $Y$, since that would mean we have a finite subcover.

We continue so on to get a sequence $y_n$ so that $d(y_k, y_j) \geq r_0 /2$ for $k > j$. This sequence is by construction not Cauchy, and no subsequence is Cauchy. So there is no convergent subsequence; indeed, this means that $Y$ is not compact.

Corollary

Let $X$ be a metric space. Take $K_1, K_2, \dots$ to be nonempty compact nesting sets, so that: $K_1 \supset K_2 \supset K_3 \supset \dots$. Then, $\bigcap\limits_{n=1}^{\infty}$ is non-empty.

Let $V_n = K_1 - K_n$ and assume that $\bigcap\limits_{n=1}^{\infty}$ is empty. Then, that means that for each $x\in K_1$, there exists some $j$ so that $x\notin K_j \implies x \in V_j$. Therefore, $V_n$ form an open cover for $K_1$.

Since $K_1$ is compact, $V_n$ have a finite subcover. Of the finite subcover, we pick the one with the smallest index $V_m$ (which necessarily contains the union of all the elements of the subcover). Then, $K_1 = V_m = K_1 - K_m$. But this means that $K_m = \emptyset$, which is a contradiction.

Finally, we conclude with some other nice properties of compact spaces.

Theorem

Let $X$ be a metric space. Then the following are true:

The first theorem tells us that in compact spaces, compactness is equivalent to being closed, whereas previously we only proved that compactness implies completeness (which is a strictly stronger condition than being closed). We can think of this as analogous to the previous theorem that in complete spaces, complete sets are compact and vice versa.

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