Connectedness And Compactness

Separation

A separation of

X

are two open and non-empty disjoint sets

A, B

such that

X = A ⊔ B

When a separation can be found for $X$ then we say that $X$ is separable.

Clopen

We say that

A \subseteq X

is clopen when

A

is open and

A

is closed

Separation Iff There Exists a Non-trivial Clopen Set

A set

X

is separable iff there exists some clopen

A \subseteq X

such that

A \neq \emptyset, X

$⟹$ Suppose that $X$ is separable, then therefore we have that $X = A ⊔ B$ where $A, B \neq \emptyset$ are both open, but then we have that $B = X ∖ A$ and since $B$ is open this shows that $A$ is closed, so it is clopen, thus we've located a non-trivial clopen set.

$⟸$ Now suppose that we have some non-trivial $A \subseteq X$ that is clopen, so then $A$ is open and closed, then $X ∖ (X ∖ A) = A$ so the set $X ∖ (X ∖ A)$ is a closed set, and therefore $X ∖ A$ is open, since $A$ is non-trivial so is $X ∖ A$ finally we note that $X = X ∖ A ⊔ A$ , thus we've constructed our separation, as needed.

Connected

We say that

X

is connected if it is non-empty and it has no separation

Connected Iff Every Clopen Is Trivial

X

is connected iff every clopen set

A \subseteq X

is one of

\emptyset, X

Suppose $X$ is connected, then if there was a clopen set that was not one of $\emptyset, X$ then there would be a separation, which is a contradiction.

Note that the above corollary is a good way to prove things. If $X$ has no clopen sets then you'd be done, otherwise given a non-empty clopen set we must prove that it must be $X$ itself, note that if the clopen set was the empty set that would be alright too. The following proposition shows how to take a non-empty clopen set and make it envelop the entire set.

A Real Interval Is Connected

[0, 1]

is connected.

Suppose that $A$ is a non-empty clopen set in $[0, 1]$ (ie, in the subspace topology) and without loss of generality assume that $0 \in A$ (which can be done because if $0 \notin A$ then $[0, 1] ∖ A$ is an open set containing $0$ ), and now lets prove that $A$ contains everything else, that is we prove: $A = [0, 1]$ .

Now let $G = {g : [0, g] \subseteq A}$ , so that our goal is to show that $1 \in G$ which would prove that $A = [0, 1]$ , since $0 \in G$ and $G$ is bounded above by $1$ then then $m = \sup (G)$ exists.

Note that $m > 0$ as $0 \in A$ and since $A$ is open then there exists some $ϵ > 0$ such that $[0, 1] \cap B (0, ϵ) = [0, ϵ) \subseteq A$ therefore $[0, \frac{ϵ}{2}] \subseteq A$ so $\frac{ϵ}{2} \in G$ therefore $m \geq \frac{ϵ}{2}$ . We will now continue by showing that $m < 1$ leads to a contradiction, and therefore we must conclude that $m \geq 1$ showing that $m = 1$ .

Since $m = \sup (G)$ then there exists a sequence of elements $g_{i} < m$ such that $g_{i} \in G$ and $g_{i} \to m$ . Since $[0, g_{i}] \subseteq A$ then $g_{i} \in A$ and since $A$ was closed, then $m \in A$ . Since $A$ is open and $m < 1$ then there exists $ϵ > 0$ such that $m + ϵ \leq 1$ and $(m - ϵ, m + ϵ) \subseteq [0, 1] = A$ but then there exists some $k \in ℕ$ such that $g_{k} > m - ϵ$ so $[0, g_{k}] \subseteq A$ so that $[0, g_{K}] \cup (m - ϵ, m + ϵ) = [0, m + ϵ] \subseteq A$ so $m + ϵ \in G$ contradictiong that $m = \sup (G)$ so we must have that $m \geq 1$ and that $m = 1$ .

A Union of Connected Sets That Intersect Is Connected

Suppose that

\forall α \in I, A_{α} \subseteq X

are connected, and that

⋂ A_{α} \neq \emptyset

then

⋃ A_{α}

is connected

Suppose that $B \subseteq ⋃ A_{α}$ is a clopen non-empty set, therefore it has some element $b$ such that there is aome $α_{0}$ such that $b \in A_{α_{0}}$ , in otherwords showing that $B \cap A_{α_{0}} \neq \emptyset$ , but $A_{α_{0}}$ is connected, therefore we have that $B \supseteq A_{α_{0}} \neq \emptyset$ or else it would have been a non trivial clopen set in $A_{α_{0}}$ .

Since we know that $B \supseteq A_{α_{0}} \supseteq ⋂ A_{α} \neq \emptyset$ , therefore we have that $\forall α, B \cap A_{α} \neq \emptyset$ and therefore for each $α \in I$ we have that $B \supseteq A_{α}$ and therefore $B \supseteq ⋃ A_{α}$ , showing that $B$ is connected.

The Continuous Image of a Connected Set Is Connected

TODO: Add the content for the proposition here.

TODO: Add the proof here.

Intermediate Value

Suppose that

X

is connected and

f : X \to ℝ

is continuous, then for any

x_{0}, x_{1} \in X

such that

f (x_{0}) < 0

and

f (x_{1}) > 0

then there is some

p \in X

such that

f (p) = 0

Note that , now suppose for the sake of contradiction taht for any $x$ we have that $f (x) \neq 0$ , so therefore $X = f^{- 1} ((- \infty, 0)) \cup f^{- 1} ((0, \infty))$ , which is a union of open sets that are non-empty so therefore $f^{- 1} ((- \infty, 0))$ and $f^{- 1} ((0, \infty))$ is a non-trivial separation of $X$

Circle Function Has a Meet Point

Let

f : S^{1} \to ℝ

be continuous, show that there is some

z \in S^{1}

such that

f (z) = f (- z)

Suppose it were not true, then if we consider the function $g (x) = f (x) - f (- x)$ , then as a difference of continuous functions then they are continuous.

Let $p \in S^{1}$ suppose it's the case that $f (p) > f (- p)$ , so then if that's the case then then we have that $g (p) > 0$ .

If $p \in S^{1}$ then also $‖ - p ‖ = | - 1 | ‖ p ‖ = 1$ therefore $- p \in S^{1}$ , then note that $g (- x) = - g (x)$ so $g (- x) < 0$ therefore since we know that $S^{1}$ is connected, then by the intermediate value theorem there exists some $z \in S^{1}$ such that $g (z) = 0$ which shows that $f (z) = f (- z)$ and therefore we have a contradition, so the statement must hold true. Note that when $f (p) < f (- p)$ we may use the same proof mirrored.

Path

A path from

a

b

(which are elements of a set

X

) is a function

γ : [0, 1] \to X

which is continuous, and we have that

γ (0) = a

and that

γ (1) = b

Path Connected

A non-empty set

X

is said to be path-connected if for any

a, b \in X

there is a path between them.

Path Connected Implies Connected

As per title.

TODO: Add the proof here.

The Topologists Sine Curve Is Connected but Not Path Connected

Consider

S = {(x, \sin (\frac{1}{x}))}

and then we have

\overline{S} = {0} \times [- 1, 1]

which we denote as the topologists sine curve, then

\overline{S}

is connected but not path connected

Firstly we note that $S$ is the image of the connected set $[0, 1)$ under a continuous map and therefore $S$ is connected, thus $\overline{S}$ is connected.

It is not path connected, for the sake of contradiction suppose that it was, then there would be path $γ : [0, 1] \to \overline{S}$ from $(0, 0)$ to any point $s \in S$ , since the set ${t : γ (t) \in 0 \times [- 1, 1]}$ is closed, then it has a largest element $b$ .

Now we know that $γ (b) \in {0} \times [- 1, 1]$ and that for any $t \in [b, 1]$ we have that $γ (t) \in S$ as $γ (t) \notin {0} \times [- 1, 1]$ . Now we can re-parameterize $γ$ on $[b, 1]$ to $γ^{'}$ defined on $[0, 1]$ .

Write $γ^{'} (t) = (γ_{1}^{'} (t), γ_{2}^{'} (t))$ , with this we have $γ^{'} (0) = (0, b)$ and that for any $t > 0$ we have $γ_{1}^{'} (t) > 0$ and that $γ_{2}^{'} (t) = \frac{1}{γ_{1}^{'} (t)}$ . For the contradiction we will contradict the continuity of $γ_{2}^{'}$ by constructing a sequence $t_{n} \to 0$ such that $γ_{2}^{'} (t_{n}) = {(- 1)}^{n}$ , that is we will construct a sequence whose image doesn't converge, showing that it is not continuous by the contra-positive.

We proceed by fixing $n \in ℕ_{1}$ and then we can pick an element $c \in (0, γ_{1}^{'} (\frac{1}{n}))$ of the form $c = \frac{1}{m π + \frac{π}{2}}$ so that $γ_{2}^{'} (c) = {(- 1)}^{n}$ (by choosing $m$ to have the same parity as $n$ ). Then we apply the intermediate value to find a $t_{n}$ such that $t_{n} \in (0, \frac{1}{n})$ and $γ_{1}^{'} (t_{n}) = c$ , thus we've constructed the contradicting sequence, so it must be impossible that $\overline{S}$ is path connected.

Note that due to the above theorem it's not true that the closure of any path connected set is always path connected, this is seen with $S = {(x, \sin (\frac{1}{x}))}$ is clearly path connected via the given map, but as noted above $\overline{S}$ is not path connected.

Balls in Rn Are Path Connected

Suppose that

x \in ℝ^{n}

and that

ϵ \in ℝ^{> 0}

then

B (x, ϵ)

is path connected

TODO: Add the proof here.

Any Open Connected Subset of Rn Is Path-connected

Suppose that

U \subseteq ℝ^{n}

is open and connected, then

U

is path connected

Let $p \in U$ and set $R_{p}$ to be the collection of points which are reachable by a path from $p$ , we'll show this set is open, so consider any $q \in R_{p}$ then since $R_{p} \subseteq U$ we know that $q \in U$ which is open and thus there exists a ball $B_{d} (q, ϵ) \subseteq U$ but now consider any $b \in B_{d} (q, ϵ)$ since $B_{d} (q, ϵ)$ is path-connected then by joining the path from $p$ to $q$ and the path frpm $q$ to $b$ we obtain a path from $p$ to $b$ so then $b$ is reachable with a path from $p$ so that $b \in R_{p}$ this shows that $q \in B (q, ϵ) \subseteq R_{p}$ so that $R_{p}$ is an open set in $U$

Following a similar thread we can actually show that $U ∖ R_{p}$ is an open set, to see why let $j \in U ∖ R_{p}$ since $j \in U$ and $U$ open then there exists $ϵ$ such that $B_{d} (j, ϵ) \subseteq U$ , now if any point in $B_{d} (j, ϵ)$ could be joined to $p$ we have a problem, as since $B_{d} (j, ϵ)$ is connected, then we can connected one path to another constructing a path from $j$ to $p$ which is impossible since $j$ was one of the elements which couldn't be reached from $p$ , this shows that $j \in B_{d} (j, ϵ) \subseteq U ∖ R_{p}$ which means that $U ∖ R_{p}$ is an open set, so we conclude that $R_{p}$ is closed.

Now we know that $R_{p}$ is clopen and that $p \in R_{p}$ so it is non-empty, since $U$ is connected, then this implies that $R_{p} = U$ to avoid contradiction, so that all points in $U$ are reachable from the point $p$ showing that $U$ is path connected (connect the two paths you obtain).

Open Cover

Given a topological space

X

and a subset

S \subseteq X

, an open cover of

S

is a collection of open sets

{U_{α}}_{α \in A}

(where

A

is an index set) such that:

S \subseteq ⋃_{α \in A} U_{α}

Subcover

Suppose that

𝒞

is a cover of

S \subseteq X

then we say that a set

𝒮 \subseteq 𝒞

is a subcover of

S

𝒮

is an open cover of it

Compact

Given a set

S \subseteq X

in a topological space

X

it is compact if every open cover of

S

has a finite subcover

The Finite Complement Topology on an Uncountable Space Is Compact

Suppose that

X

is an uncountable set with the finite complement topology, then

X

is compact.

Let ${U_{α}}_{α \in A}$ be an open cover of $X$ in the finite complement topology, so that: $X \subseteq ⋃_{α \in A} U_{α} .$ since each $U_{α}$ is open in $X$ , then complement $X ∖ U_{α}$ is finite for each $α \in A$ .

If there exists some $U_{α_{0}} = X$ , then ${U_{α_{0}}}$ covers $X$ , then that itself is an finite sub-cover of $X$ so we are done.

Now suppose that none of the $U_{α}$ are equal to $X$ , and let $β \in A$ and consider $U_{β}$ , since $X ∖ U_{β}$ is finite and non-empty so it equals ${u_{1}, \dots, u_{n}}$ , since ${U_{α}}_{α \in A}$ was an open cover, then for each $u_{i}$ there exists a $U_{α_{i}}$ containing it, thus ${U_{β}, U_{α_{1}}, \dots, U_{α_{n}}}$ is a finite subcover as needed.

The Countable Complement Toplogy on an Uncountable Set May Not Be Compact

Suppose that

X

is uncountable, then in the countable-complement topology

X

may not be compact.

We show this by constructing an explicit counter example, so we consider $X = [0, 1]$ , in this scenario then we can focus on a very specific covering by first constructing some holes $H_{n} = {\frac{1}{j} : j \geq n}$ , and then then considering $A_{n} = [0, 1] ∖ H_{n}$ , by construction the complement of these sets are countable, and thus these are open sets in $X$ , moreover the collection ${A_{n} : n \in ℕ_{1}}$ is a covering of $X$ as even for any element of the form $p = \frac{1}{i}$ for $i \in ℕ_{1}$ then $p \in A_{n + 1}$ showing that the holes are actually filled in.

Now that we've shown that the $A_{n}$ 's are an open cover, then if for the sake of contradiction our set $X$ was compact then there would exists a finite subcover $F$ , since the cover is finite there is a largest $K$ such that $A_{K} \in F$ . Take a moment to realize that $H_{i} ⊋ H_{i + 1}$ which shows that $A_{i} ⊊ A_{i + 1}$ thus there exists an element $c \in A_{K + 1} \subseteq [0, 1]$ which is not in $A_{j}$ for any $j \leq K$ and therefore $c$ is not an element of the cover, which is a contradiction because it's supposed to cover $X = [0, 1]$

Every Compact Subspace of a Hausdorff Space Is Closed

As per title.

TODO: Add the proof here.

Every Compact Subspace of a Metric Space Is Closed and Bounded

Every compact subspace

A

of a metric space

M

is closed and bounded.

Recall that any metric space is Hausdorff, and therefore

A

is closed. Now we show that it is bounded, we will use compactness for this, to this end fix

a \in A

and consider the collection of increasing open sets

{B_{d} (a, n) \cap A : n \in ℕ_{1}}

which clearly cover

A

, and thus there exists a finite subcover

F

, let

K \in ℕ_{1}

be the largest satisfying

B_{d} (a, K) \in F

, due to the fact that these sets are increasing then

B_{d} (a, K) \supseteq A

which shows that

A

must be bounded.

Note that the converse of the above is not true, for that we just have to consider an infinite space $X$ endowed with the discrete topology, in such a case $X$ is metrizable from the discrete metric $d$ , we can see the the space is bounded because $X \subseteq B_{d} (p, 1.1)$ for any $p \in X$ , moreover $X$ is closed in itself, but on the other hand if we consider the open cover which is ${{x} : x \in X}$ then any finite subcover will only have finitely many elements of $X$ and since $X$ was infinite, then we have a contradiction, so it must be the case that it is not compact.

Every Closed Subspace of a Compact Space Is Compact

TODO: Add the content for the proposition here.

TODO: Add the proof here.

The Image of a Compact Set Under a Continuous Function Is Compact

TODO: Add the content for the proposition here.

TODO: Add the proof here.

Hausdorff Spaces Separate Compact Subspaces and Points

Y

is a compact subspace of a hausdorff space

X

and

x_{0}

is not in

Y

, then there exist disjoint open sets

U

and

V

X

containing

x_{0}

and

Y

respectively.

TODO: Add the proof here.

Disjoint Compact Subspaces of a Hausdorff Space Can Be Separated

Suppose that

A, B

are disjoint compact subsets of a Hausdorff space

X

then there exists disjoint open sets

U, V

such that

A \subseteq U

and

B \subseteq V

Recall that for each $a \in A$ we obtain some disjoint open sets $U_{a}, V_{a}$ in $X$ such that $a \in U_{a}$ and $B \subseteq V_{a}$ thus the collection ${U_{a} : a \in A}$ is an open cover of $A$ and since it is compact then there exists a finite subcover ${U_{a_{i}} : 1 \leq i \leq n}$ for some $n \in ℕ_{1}$ , since each $V_{a_{i}} \supseteq B$ then we know $V = ⋂ {i = 1}^{n} V_{a_{i}} \subseteq B$ additionally as a finite intersection it is an open set.

Now if we construct $U = ⋃_{i = 1}^{n} U_{a_{i}}$ , it also also an open set containing $A$ , now we conclude by showing that they are disjoint, to that direction suppose that there was some $c \in U \cap V$ for the sake of contradiction, then we know that there exists some $a_{j}$ such that $c \in U_{a_{j}}$ but also $c \in V$ which was the intersection so also $c \in V_{a_{j}}$ but as noted earlier $U_{a_{j}} \cap V_{a_{j}} = \emptyset$ which is a contradiction, and therefore we must have $U, V$ being disjoint, as needed.

A Continuous Function From a Compact Space to a Hausdorff Space Is Closed

Suppose that

f : X \to Y

is continuous where

X

is compact and

Y

is hausdorff, then

f

is closed.

To show that

f

is closed we suppose that

C

is closed in

X

, therefore we know it is also compact as a closed subspace of a compact set, therefore we have that

f (C)

is compact in

Y

as the continuous image of compact, but now we have a compact set in a hausdorff space so thus it is closed, therefore

f

is a closed map.

There Is Always Space for an Open Rectangle Between a Cartesian Product of Compact Spaces and an Open Set Containing Them

Suppose that

X, Y

are topological faces and suppose that

A, B \subseteq X, Y

be compact spaces respectively, then suppose that

W

is an open set such that

A \times B \subseteq W \subseteq X \times Y

then there exists an open rectangle

U \times V \subseteq X \times Y

such that

A \times B \subseteq U \times V \subseteq W

Let $a \in A$ and consider ${a} \times B$ then we know that ${a} \times B \subseteq W$ as $A \times B \subseteq W$ . Since $W$ is an open set, then for each $p \in {a} \times B$ we can obtain some $J_{p} \times K_{p}$ containing $p$ and contained within $W$ , thus forming an open cover, but since ${a} \times B$ is compact, then we obtain a finite subcover of the form ${J_{p_{1}} \times K_{p_{1}}, \dots, J_{p_{N}} \times K_{p_{N}}}$ .

Now if we consider $U_{a} = ⋂_{i = 1}^{N} J_{i}$ and $V_{a} = ⋃_{i = 1}^{N} K_{i}$ , then we know that $U_{a}$ is an open neighborhood of $a$ and $V_{a}$ is an open set in $Y$ containing $B$ which is to say that ${a} \times B \subseteq U_{a} \times V_{a}$ . Moreover observe that $U_{a} \times V_{a} \subseteq W$ and also that $U_{a} \times V_{a}$ is open in $X \times Y$ so that if we iterate $a$ over all values in $A$ namely we consider the set ${U_{a} \times V_{a} : a \in A}$ then it becomes an open cover of $A \times B$ , but since $A \times B$ is compact, then there is a finite subcover for the form ${U_{a_{1}} \times V_{a_{1}}, \dots, U_{a_{l}} \times V_{a_{l}}}$ .

Now similarly we set $U = ⋃_{i = 1}^{l} U_{a_{i}}$ and $V = ⋂_{i = 1}^{l} V_{a_{i}}$ where we know that $U, V$ are open in $X, Y$ respectively, we'll try to prove that $A \times B \subseteq U \times V \subseteq W$ . So let $(x, y) \in A \times B$ then there exists some $s \in [k]$ such that $(x, y) \in U_{a_{s}} \times V_{a_{s}}$ , now as $U$ is a union of the $U_{a_{i}}$ 's then we have that $x \in U$ , on the other hand since each $V_{a_{i}}$ contains $B$ then so does the intersection, that is $B \subseteq V$ , so now since we know $y \in B$ we automatically get that $y \in V$ showing $x, y \in U \times V$ , so that $A \times B \subseteq U \times V$ .

Now if we consider $(x, y) \in U \times V$ then again there is some $l$ such that $(x, y) \in U_{a_{l}} \times V_{a_{l}}$ and then from earlier we recall that $U_{a_{l}} \times V_{a_{l}} \subseteq W$ as it is an element from the original open cover where this held true.

Properties of the Infimum Metric and Compact Spaces

Let

X

be a metric space and let

A \subseteq X

be non-empty. Let

d (x, A) = \inf_{a \in A} d (x, a)

and let

U_{ϵ} (A) = {x : d (x, A) < ϵ}

Show that $d (x, A) = 0$ iff $x \in A$
Show that if $A$ is compact then $d (x, A) = d (x, a)$ for some $a \in A$
Show that $U_{ϵ} (A)$ is the union of all the $ϵ$ -balls whose centers lie in $A$
If $A$ is compact and $U \supseteq A$ is open, show that there is some $ϵ > 0$ wherein $U_{ϵ} (A) \subseteq A$
Find a counterexample to the result in the above if $A$ is not assumed to be compact

Suppose that $d (x, A) = δ > 0$ , then $B_{d} (x, δ) \cap A = \emptyset$ so $x \notin \overline{A}$ , because we've found an open set around $x$ which doesn't intersect $A$ . For the other direction we assume that $x \notin \overline{A}$ then by the definition there is an open set containing $x$ that doesn't intersect $A$ , which is equivalent to some $δ > 0$ existing where we have $B_{d} (x, δ) \cap A = \emptyset$ so we can deduce that $d (x, A) > \frac{δ}{2} > 0$

We'll now prove that if $A$ is compact then $d (x, A) = d (x, a)$ .

Recall that we've proven the distance function continuous, given $x \in X$ define the function $d_{x} (a) = d (x, a) : A \to ℝ$ which is also continuous. Now since $A$ was a compact set then by the extreme value theorem we obtain some $m \in A$ such that $d_{x} (m)$ is the minimum value on $A$ , that is $d (x, m) \leq d (x, y)$ for any $y \in A$ in otherwords this is the infiumum so we conclude that $d (x, A) = d (x, m)$ as needed.

We now prove that $U_{ϵ} (A)$ is the union of all $ϵ$ balls whose centers lie in $A$ . Let $p \in U_{ϵ} (A)$ therefore we know that $d (p, A) = \inf {d (p, a) : a \in A} < ϵ$ , therefore there must be an $a \in A$ such that $d (p, a) < ϵ$ or else the infimum could not be less than $ϵ$ which implies that $p \in B_{d} (a, ϵ)$ , therefore $U_{ϵ} (A) \subseteq ⋃_{a \in A} B_{d} (a, ϵ)$ .

Now suppose that $x$ is in the union, ie there is some $a \in A$ such that $x \in B_{d} (a, ϵ)$ so then we know that $d (x, A) \leq d (x, a) < ϵ$ so therefore $x \in U (A, ϵ)$ showing the other inclusion, therefore $U_{ϵ} (a) = ⋃_{a \in A} B_{d} (a, ϵ)$

If $A$ is compact and $U \supseteq A$ is open, then there exists some $ϵ > 0$ such that $U_{ϵ} (A) \subseteq U$ . Since $U$ is open and $A \subseteq U$ then for any $x \in A$ we know that ther is some $ϵ_{x} > 0$ so that $B_{d} (x, ϵ_{x}) \subseteq U$ thus we may construct the open cover ${B_{d} (x, ϵ_{x}) : x \in A}$ of $A$ , since it is compact then we obtain a finite subcover of the form ${B_{d} (x_{i}, ϵ_{x_{i}}) : i \in [n]}$ . If we set $l$ to be the lebesgue number for this covering, then by considering $ϵ = \frac{l}{2}$ , then any ball of the form $B_{d} (a, ϵ)$ has a diameter less than $l$ so we obtain some $j \in [n]$ such that $B_{d} (a, ϵ) \subseteq B_{d} (x_{i}, ϵ_{x_{i}})$ so we have that $⋃_{a \in A} B_{d} (a, ϵ) \subseteq ⋃_{i = 1}^{n} B_{d} (x_{i}, ϵ_{x_{i}}) \subseteq U$ , since we just proved that $U_{ϵ} (a) = ⋃_{a \in A} B_{d} (a, ϵ)$ , then we are done.

Note that the above is not necessarily true if $A$ is not compact, as if we consider the closed set $A = {(x, \tan (x)) : x \in (\frac{- π}{2}, \frac{π}{2})}$ which fails to be compact as it is not bounded, and consider $U = (\frac{- π}{2}, \frac{π}{2}) \times ℝ$ which is open in $ℝ^{2}$ but if we consider any $ϵ > 0$ then there is a $p \in A$ such that $B_{d} (p, ϵ) \cap (ℝ^{2} ∖ U) \neq \emptyset$ , to see this for any epsilon ball as we slide up $\tan (x)$ we get aribtrariliy close to the line $x = \frac{π}{2}$ and thus we eventually break out of the box. Thus we've shown that there is no epsilon neighborhood of $A$ which is entirely contained within $U$ as needed.