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I have some problems understanding the following definition

DEFINITION

  1. A set of points A of the xy plane is called connected if any points of A can be joint by a continuous curve lies entirely in A.

  2. A set of points A of the xy plane is called open if each point of A is the center of a circle whose interior lies entirely in A.

  3. An OPEN + CONNECTED set is called a domain.

  4. A point P is called a boundary point of domain D if every circle about P contains both points in D and not in D.

  5. A DOMAIN + BOUNDARY POINTS is called closed domain.

I don't really understand the 2. and 4.

A question for 1.

Take this scenario

If we join $P$ and $Q$ by a straight line there will be a removable discontinuity between them called it $a$. If we connect $P$ and $Q$ with a curvy line such that there is no discontinuity in between then they can be joined. Is this what CONNECTED really means?

Crazy
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    "Connectedness" is actually a much deeper topological question, but in $\mathbb{R}^n$ path-connectedness (which is what you have described) is sufficient to show that a set is connected. A set is path-connected if any two points in the set can be joined by a path. This path (a continuous curve) needn't be a straight line. For example, a circle (the boundary of a disk–the interior is excluded) is path-connected, but no two points in the circle can be joined by a straight line in $\mathbb{R}^2$. – Xander Henderson Oct 05 '17 at 13:38
  • That being said, I am not sure what you are asking. You say that you are confused about points (2) and (4), but you seem to be asking a question about (1). Can you please clarify? – Xander Henderson Oct 05 '17 at 13:39

1 Answers1

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For your scenario with (1):

What do you mean by removable discontinuity in this case? The definition 1 simply says that if there exists, for each pair of points in $A$, a curve in $A$ that joins the points, then $A$ is connected. There is no "removable discontinuity" mentioned, and I don't even know how it could enter the discussion here.

For (2):

Which part do you "not understand"? What (2) says is a definition. There is nothing to understand here, you can just accept the definition, and then try some exercises to internalize the definition. By definition, a set $A$ is closed if every point $(x,y)\in A$ is the center of some circle that is in $A$. Let's take an example.

For example, if $A$ is the disc $$A=\{(x,y)| x^2+y^2<1\}$$centered at $0$ and the radius of $1$, then $A$ is open. You can see this by showing that if $(x,y)\in A$, then the circle with the center at $(x,y)$ and the radius of $\min\{1-|x|, 1-|y|\}$ is a subset of $A$.

For $4$, again, I don't know what you mean by "I don't understand", but you can see, for example, that $(1,0)$ is a boundary point of $A$ from above. You can do this by proving the statement:

If $D$ is a disc centered at $(1,0)$, then $D$ contains both points from $A$ and points not in $A$.

5xum
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