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fishturtle1

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- 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.

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.

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