Apostol definition of interior point and open set

[kahwawashay1]

S is a set such that S$\subseteq$Rⁿ
point a is in S: a$\in$S

The point "a" is an interior point of S if there is an open n-ball with center "a", all of whose points belong to S.
ie., every interior point of S can be surrounded by an n-ball such that B(a)$\subseteq$S, where B(a) is the set of all points x in Rⁿ such that ||x-a||<r (r is radius of the ball)

Ok so let's say we have such an S with point "a" and also containing all of the points in B(a).

Now, according to Apostol, the "interior" of S is the set of all interior points of S and is denoted by int S.
So let Q=int S for the S that I defined above. So, Q={a}, since "a" is the only interior point of S.

Now comes "open set"
A set W is an open set if all its points are interior points. ie., W is open set if and only if W = int W

This is what I don't understand. And actually, instead of "W", Apostol used S again, but I do not think he means any relation with the previous S's?
I know that an open set is basically an open interval, like (3, 4) on the x-axis would mean 3<x<4. So shouldn't an open set just be B(a)? Like in the case of S=(3, 4), S is an open set with interior point a=3.5, since 3.5 is the center, and B(a) is all x such that ||x-3.5||<0.5 ..but that means that S contains points other than just int S; namely, it also contains all points satisfying B(a), which are not interior points of S, so S≠int S


?

 

[mathwonk]

you did not define a set S. If you mean the statement that S is a set containing a ball B(a), although this does not define a set, still it implies that any such S has as interior points all of B(a).

going out on a limb here:
you are either just getting started or in over your head in apostol, or have a teacher who does not explain much.

 

[Dr. Seafood]

Hopefully this will clear some stuff up.

Let $S \subseteq \mathbb{R}^n$. A point [itex]x \in S[\itex] is called an interior point of S if there exists a number r > 0 such that, whenever $a \in \mathbb{R}^n$ is such that $\Vert a - x \Vert < r$, we have $a \in S$.

To internalize this definition, let $D_r(x) = \{a \in \mathbb{R}^n : \Vert a - x \Vert < r\}$ denote the collection of points within a fixed radius "r" (> 0) of the point "x" in S. We can call sets of this form open neighborhoods (of x). In $\mathbb{R}^2$ this is the collection of points in the disk centered at x with radius r (but not on the circle). D_r(x) is the set of points "within r" from x. Of course $x \in D_r(x)$ for any r > 0.

If I can draw a circle around a point $x \in \mathbb{R}^2$ such that the entire circle is contained in S, then x is an interior point of S. In other words, an interior point of a set is a point around which I can draw a circle (of some arbitrary radius) completely contained in the set. Notice that a point outside my set can't have this property. Also notice that a point on the "edge" of my set can't be an interior point -- it seems that any circle centered at a point on the boundary would have a part lying outside the set. Draw a picture to see this! So when I say "interior point", I mean a point truly on the "inside" (or "interior") of the set.

Notice that the definition of interior point of a set S is then $x \in \mathbb{R}^n : (\exists r > 0 : D_r(x) \subseteq S)$. If every point in a set is an interior point of that set, then we call the set open.

Now let the interior of S be denoted $\mathrm{int} S = \{x \in \mathbb{R}^n : (\exists r > 0 : D_r(x) \subseteq S)\}$. This is simply the collection of interior points of S. Now an open set is a set S for which S = int S. This means that every point in S is an interior point of S.

In $\mathbb{R}^n$, open sets are precisely those which are unions of open neighborhoods.

 

[SteveL27]

I know that an open set is basically an open interval, like (3, 4) on the x-axis would mean 3<x<4. So shouldn't an open set just be B(a)?

Little more to it than that. For example consider the union of (1,2) and (3,4). Is this an open set?

How about the union of (1,2), (3,4), (5,6), (7,8), ... which is a union of infinitely many disjoint intervals. Does it fit the definition of an open set?

Do you see how Apostol's definition covers these cases as well as the case of a simple interval?

 

[deluks917]

You are misunderstanding int(S). Int(S) usually has more then one member. For example S = [0,10]. Then 1,2,3,7.8,pi etc are all in int(S). In fact int(S) = (0,10).

 

[kahwawashay1]

Hopefully this will clear some stuff up.

Let $S \subseteq \mathbb{R}^n$. A point [itex]x \in S[\itex] is called an interior point of S if there exists a number r > 0 such that, whenever $a \in \mathbb{R}^n$ is such that $\Vert a - x \Vert < r$, we have $a \in S$.

To internalize this definition, let $D_r(x) = \{a \in \mathbb{R}^n : \Vert a - x \Vert < r\}$ denote the collection of points within a fixed radius "r" (> 0) of the point "x" in S. We can call sets of this form open neighborhoods (of x). In $\mathbb{R}^2$ this is the collection of points in the disk centered at x with radius r (but not on the circle). D_r(x) is the set of points "within r" from x. Of course $x \in D_r(x)$ for any r > 0.

If I can draw a circle around a point $x \in \mathbb{R}^2$ such that the entire circle is contained in S, then x is an interior point of S. In other words, an interior point of a set is a point around which I can draw a circle (of some arbitrary radius) completely contained in the set. Notice that a point outside my set can't have this property. Also notice that a point on the "edge" of my set can't be an interior point -- it seems that any circle centered at a point on the boundary would have a part lying outside the set. Draw a picture to see this! So when I say "interior point", I mean a point truly on the "inside" (or "interior") of the set.

Notice that the definition of interior point of a set S is then $x \in \mathbb{R}^n : (\exists r > 0 : D_r(x) \subseteq S)$. If every point in a set is an interior point of that set, then we call the set open.

Now let the interior of S be denoted $\mathrm{int} S = \{x \in \mathbb{R}^n : (\exists r > 0 : D_r(x) \subseteq S)\}$. This is simply the collection of interior points of S. Now an open set is a set S for which S = int S. This means that every point in S is an interior point of S.

In $\mathbb{R}^n$, open sets are precisely those which are unions of open neighborhoods.

thankss i get it now! :D

 

[kahwawashay1]

going out on a limb here:
you are either just getting started or in over your head in apostol, or have a teacher who does not explain much.

idk my class is called calculus I and II. but my teacher does more analysis stuff. not as deep as actual analysis tho, so i guess i am a bit in over my head lol

 

[diligence]

idk my class is called calculus I and II. but my teacher does more analysis stuff. not as deep as actual analysis tho, so i guess i am a bit in over my head lol

you may be in over your head now, but starting out with this level of rigor will be incredibly amazing for you in the future, assuming you continue in mathematics.