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Can you explain me the definition of a boundary point ?

The definition is :

Let $A \subset \mathbb{R}^{n}$. A point $x \in \mathbb{R}^{n}$ is called a boundary point of $A$ if every neighborhood of $x$ contains at least one point in $A$ and a least one point not in $A$.

I attach a draw. Boundary point

Following the definition all the pictures are correctly but in my opinion I think that the last picture is OK when the boundary point is laying on that edge.

Another example, let's consider the following example :

Let $A=(a,b) \subset \mathbb{R}$. Then the boundary points of $A$ consists of te points $a$ and $b$. Why only the points $a$ and $b$? Why not the point $b+1$ or $a+1$ ?

Thanks :)

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What is a neighborhood of a point? –  Babak S. Aug 24 '12 at 13:59
1  
@BabakSorouh mathworld.wolfram.com/Neighborhood.html –  Iuli Aug 24 '12 at 14:02

4 Answers 4

up vote 8 down vote accepted

Your first two pictures aren’t really helpful, so I’ve made better versions:

enter image description here

In the first picture $V$ is a neighborhood of the red point that does not contain any point not in $A$, so the red point is not a boundary point of $A$. In the second picture $V$ is a neighborhood of the red point that does not contain any point of $A$, so again the red point cannot be a boundary point of $A$. Only in your third picture is it true that every neighborhood of the red point must contain points of $A$ and points not in $A$, so it’s the only picture in which the red point is a boundary point of $A$.

The point $b+1$ is not a boundary point of $(a,b)$ because it has a neighborhood that does not contain any point of $(a,b)$. In fact it has many such neighborhoods, but one easy one is $\left(b+\frac12,b+2\right)$: $b+1\in\left(b+\frac12,b+2\right)$, but $\left(b+\frac12,b+2\right)\cap(a,b)=\varnothing$.

If $b=a+1$, then of course $a+1$ is a boundary point of $(a,b)$: every neighborhood of $b$ contains points less than $b$ that are in $(a,b)$ and points bigger than $b$ that are not in $(a,b)$. If $a+1<b$, then $a+1\in(a,b)$, so $(a,b)$ itself is a neighborhood of $a+1$ that contains no points of $\Bbb R\setminus(a,b)$; this shows that $a+1$ is not a boundary point of $(a,b)$ in this case. If $a+1>b$, then $(b,a+2)$ is a neighborhood of $a+1$ that contains no points of $(a,b)$, and again $a+1$ is not a boundary point of $(a,b)$. (You’ll probably find it helpful to make drawings of these different cases.)

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In your $A=(a,b)\subset\mathbb R$, $a$ and $b$ is the only boundary points as you noted because: If you equipped $a$ (or $b$) with any ball centered at $a$ d with any radius, you would have infinite points in an out of $A$ in intersection. Indeed, $B_x\cap A\neq1,B_x\cap\mathbb R\neq1, $ –  Babak S. Aug 24 '12 at 14:19

I think the key word in the definition that you may be overlooking is every neighborhood. Just because you can find a single neighborhood that contains points both inside and outside the set does not mean it is a boundary point.

For example, for $(a,b)$ the point $b+1$ is not a boundary point because $((b+1)-1/2, (b+1)+1/2)=(b+1/2, b+3/2)$ is a neighborhood of $b+1$ that contains no point of $(a,b)$.

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Following the definition not all the pictures are correct. The definition says that every neighborhood of $ x $ contain at least one point in $A$ and at least one point not in $A$. In the upper pictures you can find a neighborhood such that there are no points in $A$ or all the points are in $A$ respectively

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The key word is "every" ! In your last example, the point $a-1$ has a neighborhood consisting of only points outside of $A$ : $(a-3/2 , a-1/2)$.

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