## Metric Space

• Simplest mathematical structure. We need to talk about “distance,” which allows us to talk about “$$\epsilon$$-neighborhoods” and “convergence”, etc.

Definition: A metric space $$(S, d)$$ is a set $$S$$, and a distance function $$d: S \times S \rightarrow \mathbb{R}$$ that satisfies:

1. $$d(x,x) = 0\ \forall x \in S$$
2. $$d(x, y) > 0\ \forall x, y \in S \textrm{ and } x \neq y$$
3. $$d(x,y) + d(y,z) \geq d(x,z)\ \forall x,y,z \in S$$
4. $$d(x,y) = d(y,x)$$

### Examples

• $$S = \mathbb{R}^2, \textrm{ and } d((x_1, y_1), (x_2, y_2)) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$
• $$S = \mathbb{R}^2, \textrm{ and } d((x_1, y_1), (x_2, y_2)) = 0 if (x_1, y_1) = (x_2, y_2), 1 if (x_1, y_1) \neq (x_2, y_2)$$
• $$S = \mathbb{R}, d(x, y) = \vert x-y \vert$$

## Neighborhoods

Definition: Let $$(S, d)$$ be a metric space, $$x \in S, r > 0$$. We can create a Ball with radius $$r$$ that contains all elements within that radius, denoted $$B_r(x) := \{y \in S \vert d(x, y) < r\}$$.

### Examples

$$S = \mathbb{R}^2, d((x_1, y_1), (x_2, y_2)) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$.
$$B_1(0, 0) =$$ circle with radius 1, boundary excluded.

$$S = \mathbb{R}, d(x, y) = \vert x-y \vert$$
$$B_1(1) = (0, 2)$$

## Open Subsets

Definition: Let $$(S, d)$$ be a metric space, and $$E \subset S$$. We say $$E$$ is an open subset of $$S$$ if $$\forall x \in E, \exists r > 0$$ such that $$B_r(x) \in E$$.

### Examples

Q: Let $$(R, d)$$ be a metric space. Is $$[0, 1] \subset \mathbb{R}$$ an open subset?

Answer: No. If we take a look at the “edge” (ie. point 0), we can see that the ball around this point is not in the set. Formally, $$\forall r > 0, B_r(1) \not\subset [0, 1]$$ since $$B_r(1)$$ implies the range $$(1-r, 1+r)$$, which contains elements outside.

Q: Taking the same setup above, is $$(0, 1) \in \mathbb{R}$$ open?

Answer: Any element must be in $$(0, 1)$$, so we can always take the radius to be half the distance to the shorter border.
Formally, if $$r = min\{x, 1-x\}$$, then $$B_r(x) \subset (0, 1)$$.

Q: In general, for any metric space, is $$B_r(x) \in S$$ open?

To prove this, we must show $$\forall y \in B_r(x)$$, we have some $$r' > 0$$ such that $$B_{r'}(y) \subset B_r(x)$$.
Let us define $$d(x, y) = r_0$$.
Claim: $$B_{r - r_0}(y) \in B_r(x)$$.
$$\forall z \in B_{r - r_0}(y), d(z, y) < r - r_0$$. By Triangle Inequality, $$d(z, x) \leq d(z, y) + d(y, x) < (r - r_0) + r_0 = r$$. This means that $$z \in B_r(x)$$, which completes the proof.

## Closed Subsets

Definition: Let $$(S, d)$$ be a metric space with $$E \in S$$. We say $$E$$ is a closed subset of $$S$$ if $$E^C \in S$$ is open (where $$E^C = \{x \in S \vert x \not\in E\})$$.

### Examples

Q: Is $$[0, 1] \in R$$ closed?

Yes. $$[0, 1]^\complement = (-\infty, 0) \cup (1, \infty)$$. We can then use the same argument as above to show that it is open.

Q: Is $$(0, 1) \in R$$ closed?

No. $$(0, 1)^\complement = (-\infty, 0] \cup [1, \infty)$$. We can then use the same argument as above to show that it is not open.

Q: Is $$E := \{\frac{1}{n} \vert n \in \mathbb{R}\} \in R$$ closed?

No. $$E^\complement = (1, \infty) \cup (1/2, 1) \cup (1/3, 1/2) \cup ... \cup (-\infty, 0]$$. For any radius $$r$$ around $$0$$, we can find a $$1/n$$ such that it’s within this radius.
Equivalently, $$\forall r>0, B_r(0)$$ always contains some elements in $$E$$. This implies that $$B_r(0) \not\subset E^\complement$$, so $$E$$ is not closed.

These examples illustrate that for a set to be closed, it should contain all its limit points.

## Limit Points

Definition: Let $$(S, d)$$ be a metric space, with $$E \in S$$. We call $$x \in S$$ a limit point of $$E$$ if $$\forall r > 0, B_r(x) \cap E$$ contains a point $$y$$ such that $$y \neq x$$.

### Examples

Q: What are the limit points of $$E := (0, 1) \in R$$?

$$0$$ is a limit point, since there exists points in $$B_r(0) \not\in R\ \forall r$$ Every point in $$[0, 1]$$ is a limit point of $$(0, 1)$$.

Q: Is it true that any point $$x \in E$$ is a limit point of $$E$$?

No. Take the example $$E = \{1, 2, 3\}$$. $$1$$ is not a limit point of $$E$$.

## Closure

Definition: The closure of $$E \in S$$ is defined to be the union of $$E$$ and its limit points.

$$\overline{E} = E \cup \{\textrm{ limit points of }E\}$$
e.g. $$\overline{(0, 1)} = [0, 1]$$

Theorem: Let $$(S, d)$$ be a metric space, and $$E \subset S$$. We claim that $$E$$ is closed if and only if $$E$$ contains all the limit points of $$E$$.

Proof $$\rightarrow$$: We want to show that if $$x \not\in E$$, then $$x$$ is not a limit point of $$E$$.
$$x \not\in E$$ if and only if $$x \in E^\complement$$ is open. This implies that $$\exists r > 0$$ such that $$B_r(x) \subset E^\complement$$, which means that $$B_r(x) \cap E = \emptyset$$. This shows that x is not a limit point of $$E$$.

Proof $$\leftarrow$$: We want to show $$E^\complement$$ is open. $$\forall x \in E^\complement$$, we know $$x \not\in E = \overline{E}$$, which means $$x$$ is not a limit point of $$E$$. This means $$\exists r > 0$$ such that $$B_r(x) \cap E$$ is either \emptyset or $$\{x\}$$. The second case cannot be true, since $$x \not\in E$$. Therefore, $$B_r(x) \cap E = \emptyset$$, so $$B_r(x) \in E^\complement$$. This shows that $$E^\complement$$ is open, so $$E$$ is closed.

Exercise: Is $$\mathbb{Q} \in \mathbb{R}$$ open or closed? What about $$\overline{\mathbb{Q}}$$?