Is the set open, closed, neither, bounded, connected?

[fishturtle1]

Homework Statement

Sketch the sets defined by the following constraints and determine whether they are open, closed, or neither; bounded; connected.

Relevant Equations

Suppose $G$ is a subset of $\mathbb{C}$.

A point $a$ in $G$ is an interior point if there exists an open disk with center $a$ that is a subset of $G$.

A point in $a$ in $G$ is a boundary point if every open disk centered at $b$ contains a point in $G$ and a point not in $G$.

A set is open if it contains all only its interior points.

A set is closed if it contains all its boundary points.

The set $G$ is bounded if $G \subset D[0,r]$ for some $r$.

Two sets $X, Y$ are separated if there are disjoint open sets $A,B \subset \mathbb{C}$ so that $X \subset A$ and $Y \subset B$. A set $G$ is connected if its impossible to find two separated nonempty sets whose union is $G$. A region is a connected open set.

Let $z = a + bi$. Using the definition of modulus, we have $\vert z - 3 \vert < 2$ is equivalent to $\sqrt{(a+3)^2 + b^2} < 2$. Squaring both sides we get $(a+3)^2 + b^2 < 4$. This is the open disk center at $3$ with radius $4$ which we write as $D[-3, 2]$.

First we show $D[-3,2]$ is open.

Proof: Let $z \in D[-3, 2]$. Then $\vert -3 - z \vert < 2$. Let $\varepsilon = \frac{2 - \vert z + 3 \vert}{2}$. We will show $D[z, \varepsilon]$ is a subset of $D[-3, 2]$. Let $z_0 \in D[z, \epsilon]$. Then $\vert -3 - z_0 \vert \le \vert -3 - z \vert + \vert z - z_0 \vert < \vert -3 - z \vert + \frac{2 - \vert z + 3\vert}{2} = \frac{2\vert -3 - z \vert + 2 - \vert z +3 \vert}{2} = \frac{\vert z + 3\vert + 2}{2} < \frac{2 + 2}{2} = 2$ Thus, we have $\vert z_0 - 3 \vert < 2$. This implies $z_0 \in D[-3, 2]$. This implies $D[z, \varepsilon] \subset D[-3,2]$. Thus, $z$ is an interior point in $D[-3,2]$. Thus every point in $D[-3,2]$ is an interior point.

Next suppose $w$ is a point not in $D[-3,2]$. Let $r \in \mathbb{R}$ and consider $D[w, r]$. Since $\vert w + 3 \vert \ge 2$ and $w \in D[w, r]$, we have $D[w, r]$ is not a subset of $D[-3,2]$. We may conclude any point not in $D[-3,2]$ is not an interior point.

We may conclude $D[-3,2]$ contains all its interior points and is thus open. [] Could we have just shown that any point not in $D[-3,2]$ is not an interior point, therefore any interior point must be in $D[-3,2]$ and so $D[-3,2]$ is open?

Next we show $D[-3,2]$ is not closed.
Proof: Consider any open disk centered at $-1$, say $D[-1, r]$. Observe $-1 \in D[-1, r]$ and $-1 \not\in D[-3,2]$. Moreover, $\vert - 1 - (-1 - \frac r2) \vert = \frac r2 < r$. So $(-1 - \frac r2) \in D[-1, r]$. But $\vert -3 - (-1 - \frac r2) \vert = \vert -2 + \frac r2 \vert$ < 2. So $(-1 - \frac r2) \in D[-3, 2]$. Since $r$ was arbitrary, we have $w$ is a boundary point of $D[-3,2]$. Since $w \not\in D[-3,2]$, we conclude that $D[-3,2]$ is not closed. []It is clear that $D[-3,2] \subset D[-3, 3]$. So $D[-3,2]$ is bounded.For connectedness I'm not sure.. I take a guess that its connected and try by contradiction: Suppose there exists separated sets $X, Y$ such that $X \cup Y = D[-2,3]$. Then there is disjoint sets $A, B \subset \mathbb{C}$ such that $X \subset A$ and $Y \subset B$. So $D[-2,3] \subset A \cup B$...I want to make sure the proofs for open, closed were correct, and also I am not sure how to do the connected part.

 

[fresh_42]

A point in $a$ in $G$ is a boundary point if every open disk centered at $b$ contains a point in $G$ and a point not in $G$.

I assume this is a typo: centered at $a$!

A set is open if it contains all only its interior points.

This is wrong. A set is open if all its points are interior points. Closures of open sets also contain all interior points!

Open disks are usually written $U_r(c)$ or $U(c;r)$ and closed ones with a $B$ for ball. Your notation $D[-3,2]$ for $U_2((-3,0))$ is a bit confusing, especially as you wrote center $3$ instead of $-3$ and radius $4$ instead of $2$; and $[.,.]$ is associated with closed, not open. You wrote $|z-3|<4$ which is $|(a-3)^2+ib|<4$ and not $a+3$.

 

[fishturtle1]

I assume this is a typo: centered at $a$!

This is wrong. A set is open if all its points are interior points. Closures of open sets also contain all interior points!

Open disks are usually written $U_r(c)$ or $U(c;r)$ and closed ones with a $B$ for ball. Your notation $D[-3,2]$ for $U_2((-3,0))$ is a bit confusing, especially as you wrote center $3$ instead of $-3$ and radius $4$ instead of $2$; and $[.,.]$ is associated with closed, not open. You wrote $|z-3|<4$ which is $|(a-3)^2+ib|<4$ and not $a+3$.

Thanks for the reply and sorry for all the typos! Also, in the OP Problem statement, it should have included the constraint "$\vert z + 3 \vert < 2$". Also, the book's definition for open set is exactly what you gave. I must have misread.

Here's the corrected version:

Let $z = a + bi$, then we have $\vert z + 3 \vert = \sqrt{(a+3)^2 + b^2} < 2$. Squaring both sides gives $(a+3)^2 + b^2 < 4$. This gives the open disk of radius $2$ centered at $-3$ which we write as $U(-3; 2)$.

First we show $U(-3, 2)$ is open.
Proof: Let $z \in U(-3, 2)$. Let $\varepsilon = \frac{2 + \vert z + 3 \vert}{2}$ and consider $U(z, \varepsilon)$. Let $z_0 \in U(z, \varepsilon)$. Observe,
$$\vert z_0 - (-3) \vert \le \vert z + 3 \vert + \vert z - z_0 \vert < \vert z + 3 \vert + \frac{2 - \vert z + 3 \vert}{2} = \frac{\vert z + 3 \vert + 2}{2} < \frac{2 + 2}{2} = 2$$

Thus $\vert z_0 + 3 \vert < 2$ which implies $z_0 \in U(-3, 2)$. This implies $U(z, \varepsilon) \subset U(-3, 2)$. We may conclude every point of $U(-3, 2)$ is an interior point and therefore $U(-3, 2)$ is open. []

Next we will show $U(-3, 2)$ is not closed.
Proof: $U(-3, 2)$ is closed if it contains all its boundary points. Consider the open disk $U(-1, r)$ for some $r \in \mathbb{R}$. Then $-1 \in U(-1, r)$ and $-1 \not\in U(-3, 2)$. Moreover $\vert -3 - (-1 - \frac r2) \vert = \vert -2 + \frac r2 \vert < 2$. So $(-1 - \frac r2) \in U(-3, 2)$ and $(-1 - \frac r2) \in U(-1, r)$.

Thus $U(-1, r)$ contains a point in $U(-3, 2)$ and a point not in $U(-3, 2)$. Since $r$ was arbitrary, we have $-1$ is a boundary point of $U(-3, 2)$. Since $-1 \not\in U(-3,2)$, we have $U(-3, 2)$ it not closed. []

Lastly, $U(-3, 2) \subset U(-3, 3)$. Therefore $U(-3, 2)$ is bounded.

 

[fresh_42]

In the first part you need to define $\varepsilon$ with a minus sign in the numerator.

In the second part you used $0<r<8$, which is not arbitrary, and not as you said any $r\in \mathbb{R}$. So you have to say it the first time $r$ appears. The conclusion still works, as you only need $r$ arbitrary close to zero, which it is allowed to be. For "any open disc" to hold as said in the definition of boundary points, it is sufficient to just say: In case $r\geq 8$ we have a proper subset $U(-3,2) \subsetneq U(-1,r)$.

Open discs are also connected. Shouldn't you have proven this as well?

 

[fishturtle1]

In the first part you need to define $\varepsilon$ with a minus sign in the numerator.

In the second part you used $0<r<8$, which is not arbitrary, and not as you said any $r\in \mathbb{R}$. So you have to say it the first time $r$ appears. The conclusion still works, as you only need $r$ arbitrary close to zero, which it is allowed to be. For "any open disc" to hold as said in the definition of boundary points, it is sufficient to just say: In case $r\geq 8$ we have a proper subset $U(-3,2) \subsetneq U(-1,r)$.

Open discs are also connected. Shouldn't you have proven this as well?

I see now I should have had $\varepsilon = \frac{2 - \vert z - 3\vert}{2}$ in the first part.

For the second part I understand what you mean, since if $r = 10$, for example, then $\vert -3 -(-1 - \frac r2) \vert = \vert -2 + 5 \vert = 3 \ge 2$, so in this case $(-1 - \frac r2) \not \in U(-3, 2)$. So we'd have to consider something like $(-1 - \frac r3)$. But I think I understand what you mean by it is sufficient to show for $r$ arbitrarily close to 0.

For the connected part:

Assume by contradiction $U(-3, 2)$ is not connected. Then there exists $X, Y$ nonempty subsets of $U(-3, 2)$ such that $X \cup Y = U(-3, 2)$ where $X, Y$ are separated. By definition of separated, there exists disjoint open sets $A, B \subset \mathbb{C}$ such that $X \subseteq A$ and $Y \subseteq B$. Since $A,B$ are disjoint, so are $X, Y$. So, for any $x \in X$ and $y \in Y$, there exists $r \in \mathbb{R}$ such that $\vert x - y \vert \ge r$. I think this implies there must be some $z \in U(-3, 2)$ such that $z \not\in X$ and $z \not\in Y$. We also know $x$ is an interior point of $X$ and $y$ is an interior point of $Y$, but I'm not sure how to continue.

 

[fresh_42]

$\frac{r}{3}$ doesn't solve the problem, it only shifts it. One can simply distinguish two cases: $r\geq 8$ which trivially satisfies the condition for boundary points, and $r < 8$ with what you have written; hence all $r$.

The proof for connectedness I know uses the theorem, that the empty set and the entire space are the only subsets of a connected space, which are open and closed - and vice versa. Then as separated sets $X,Y$ are both open and closed in $X\cup Y = U(c,r)$.

For both proofs - the criterion of connectedness and the property of separated sets - one needs some basic topology, which I don't know whether you already have had it.

Another idea is to use convexity of $U(c,r)$. A path from $x\in X$ to $y\in Y$ is completely in $U(c,r)$. But as $A$ and $B$ are both open, there has to be a boundary point of them on this path which isn't part of neither $A$ nor $B$. The convexity is easy to see (use the path $\gamma\, : \,[0,1] \longmapsto t\cdot x+ (1-t)\cdot y$ and the triangle inequality), so you only have to prove the existence of this boundary point which you called $z$, and which is simultaneously not an element of $U(c,r)\subseteq A \cup B$ and an element of it as part of the path. My guess is that continuity of $\gamma$ would do it. Maybe we need to prove that $[0,1]$ is connected first.

 

[fishturtle1]

$\frac{r}{3}$ doesn't solve the problem, it only shifts it. One can simply distinguish two cases: $r\geq 8$ which trivially satisfies the condition for boundary points, and $r < 8$ with what you have written; hence all $r$.

The proof for connectedness I know uses the theorem, that the empty set and the entire space are the only subsets of a connected space, which are open and closed - and vice versa. Then as separated sets $X,Y$ are both open and closed in $X\cup Y = U(c,r)$.

For both proofs - the criterion of connectedness and the property of separated sets - one needs some basic topology, which I don't know whether you already have had it.

Another idea is to use convexity of $U(c,r)$. A path from $x\in X$ to $y\in Y$ is completely in $U(c,r)$. But as $A$ and $B$ are both open, there has to be a boundary point of them on this path which isn't part of neither $A$ nor $B$. The convexity is easy to see (use the path $\gamma\, : \,[0,1] \longmapsto t\cdot x+ (1-t)\cdot y$ and the triangle inequality), so you only have to prove the existence of this boundary point which you called $z$, and which is simultaneously not an element of $U(c,r)\subseteq A \cup B$ and an element of it as part of the path. My guess is that continuity of $\gamma$ would do it. Maybe we need to prove that $[0,1]$ is connected first.

Thank you for the reply. I haven't taken topology yet so I will try to use your second suggestion.

To show $[0,1]$ is connected:

Proof: Assume by contradiction $[0,1]$ is not connected. Then there exists nonempty separated $X, Y \subseteq [0,1]$ such that $X \cup Y = [0,1]$. By definition of separated, there exists disjoint open sets $A, B \subset \mathbb{C}$ such that $X \subseteq A$ and $Y \subseteq B$. This implies $X, Y$ are disjoint and both open. So, WLOG, $X = [0, a)$ and $Y = (b, 1]$ for some real numbers a, b where $a < b$. But can we really say this? Because now we can see X contains a boundary point, namely $0$, so $X$ is not open, a contradiction. So we can conclude $[0,1]$ is connected. []

Next we need to show $\gamma\, : \,[0,1] \longmapsto t\cdot x+ (1-t)\cdot y$ is continuous. i.e. For all $z_0 \in [0,1]$ we have $\lim_{z\rightarrow z_0} \gamma(z) = \gamma(z_0)$.

Proof: Let $\varepsilon > 0$ and $x, y \in U(-3, 2)$. Choose $\delta := \delta(\varepsilon, x, y) = \frac{\varepsilon}{2max\lbrace \vert x \vert , \vert y \vert \rbrace}$. Let $z_0 \in [0,1]$ and supposé $\vert z - z_0 \vert < \delta$.

Observe,

$$\vert \gamma(z) - \gamma(z_0)\vert = \vert zx + (1-z)y - (z_0x + (1-z_0)y)\vert = \vert zx + y - zy - z_0x - y + z_0y \vert = \vert x(z - z_0) + y(z_0 - z) \vert \le \vert x(z - z_0) \vert + \vert y(z_0 - z) \vert < \vert x \vert \varepsilon + \vert y \vert \varepsilon \le \frac \varepsilon 2 + \frac \varepsilon 2= \varepsilon$$

Thus, $\vert \gamma(z) - \gamma(z_0)\vert < \varepsilon$. We may conclude $\gamma$ is continuous on $[0,1]$. []

But I'm not sure how this will lead to a boundary point. Does it have to do with the convexity part?

 

[fresh_42]

But can we really say this?

Well, we have to use something! You can go the way with the open and closed sets by many tiny steps, which include things like "the complement of an open set is a closed set" etc. But without it, what shall we use?

More rigorously of course would be, that $A \subseteq [0,1]$ open only means, that $A$ is a union of open sets of the form $[0,a)$ and $(b,a)$. It does not necessarily have to be $[0,a)$. E.g. $$[0,\frac{1}{10}) \cup (\frac{3}{10},\frac{2}{5}) \cup \left( \cup_{n\in \mathbb{N}} \left( \frac{1}{2},\frac{2}{3}-\frac{1}{n+
15}\right)\right)$$ is an open subset of $[0,1]$, too.

Here I think we can work with two open intervals as you did. The boundary points can directly be named:
Let $X\subseteq A = [0,a+\varepsilon)$ and $Y\subseteq B=(b-\varepsilon,1]$ then the boundaries are $a+\varepsilon$ and $b-\varepsilon$. They are in $[0,1]$ but neither in $A$ nor $B$. Hence $a<b$ cannot be; or $X=\emptyset$ and $Y=[0,1]$ or vice versa. (Take $\varepsilon= (b-a)/4$ or so.)

But I'm not sure how this will lead to a boundary point. Does it have to do with the convexity part?

No, convexity is needed for the fact that our entire line $\gamma(t)$ between $x\in X$ and $y\in Y$ is within our open disc.

Do you know the in my opinion real definition of continuity? The one you used is the analytical one which works in spaces, whose topology is induced by the metric, here the Euclidean distance. However, continuous functions are the morphisms in the category of topological spaces. This means, we do not need a metric. Instead the definition goes:
A function is continuous, if the pre-image of an open set is open.
$$
f\, : \,(X,\mathcal{X}) \longrightarrow (Y,\mathcal{Y}) \text{ is continuous } \Longleftrightarrow \left( \,\forall\, O \in \mathcal{Y} \Longrightarrow f^{-1}(O) \in \mathcal{X}\right)
$$
I say this not only because it is the natural definition of continuity in topological spaces, it also immediately helps us here: you can consider $\gamma^{-1}(A)$ and $\gamma^{-1}(B)$ and get open sets in $[0,1]$.

That the two definitions are equivalent for metric spaces is not difficult to see: one open set is the $\varepsilon-$ neighborhood $|f(x)-f(x_0)|< \varepsilon$ and the other the $\delta -$ neighborhood $|x-x_0|<\delta$. Again one needs that the open sets are open intervals (or open discs, or open balls, whatever).

Edit: Another possibility is to take the same proof for the connectedness of intervals and generalize it for the complex plane. The idea is the same, only the general sets are a bit mean. Continuity allows us to pull back the situation into the real case of intervals.