Chapter 7.2, Problem 14E

(a)

To determine

To Find: The boundary points of $(2,5],(0,1),[3,5]\cup \left\{6\right\}$ and $\varnothing$ .

Answer to Problem 14E

The boundary points of the sets ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ are 3,5 and 6.

Explanation of Solution

Given Information:

The point $x$ is a boundary point of set $A$ if for all values of $\delta >0$ , ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ .

Considering $A=(2,5]$ and $x$ is a boundary point of $A$ then for all values of $\delta >0$ , ${\rm N}(x,\delta )\cap A\ne \varnothing$ therefore, $x=[2,5]$ .

Again since ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ there will be only two such points which are 2 and 5.

If we consider for $(0,1)$ similarly the boundary points we get is 0 and 1. For $[3,5]\cup \left\{6\right\}$ then 3 and 5 are the boundary points.

For $\delta >0$ we consider neighborhood $N(6,\delta )$ . This will contain an element from the set which is 6 and it also contain an element from the complement set.

Therefore, 6 is also boundary point.

Thus 3,5 and 6 are the boundary points.

Hence, the boundary points of the sets ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ are 3,5 and 6.

(b)

To determine

To Prove: If $x$ is not an interior point of $A$ and not an interior point of $A^c$ then $x$ is a boundary point of $A$ .

Explanation of Solution

Given Information:

The point $x$ is a boundary point of set $A$ if for all values of $\delta >0$ , ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ .

Prove:

Considering $x$ as a boundary point of $A$ .

If $x$ will be an interior point then $\delta >0$ so that ${\rm N}(x,\delta )\subseteq A$ . Similarly since $A\cap A^c=\varnothing$ and ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ then $x$ will contradict as a boundary point of $A$ .

Again, if $x$ is an interior point of $A^c$ then also $\delta >0$ so that ${\rm N}(x,\delta )\subseteq A^c$ which says ${\rm N}(x,\delta )\cap A\ne \varnothing$ and again $x$ will contradict as a boundary point.

Thus $x$ will not be an interior point of $A$ or $A^c$ . Then ${\rm N}(x,\delta )$ are not contained in either $A$ or $A^c$ . Therefore ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\subseteq A^c\ne \varnothing$ .

Thus, it shows that $x$ is interior point of $A$ .

Hence, proved.

(c)

To determine

To Prove: If $A$ contains none of the boundary points then $A$ is open.

Explanation of Solution

Given Information:

The point $x$ is a boundary point of set $A$ if for all values of $\delta >0$ , ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ .

Prove:

Considering $x$ to be open, so $A$ is an interior point. $A$ doesn’t contain it as boundary point is not interior point.

Alternatively, consider $A$ do not contain its boundary point.

Consider, $x\in A$ then $\delta >0$ so that ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\subseteq A^c\ne \varnothing$ . Thereby $x\in A$ so $A^c=\varnothing$ . Thus ${\rm N}(x,\delta )\subseteq A$ which shows that $x$ is an interior point and as $x$ is random so every point is interior point.

Thus, $A$ is open set.

Hence, proved.

(d)

To determine

To Prove: If $A$ contains all the boundary points then $A$ is closed.

Explanation of Solution

Given Information:

The point $x$ is a boundary point of set $A$ if for all values of $\delta >0$ , ${\rm N}(x,\delta )\cap A\ne \varnothing$ and ${\rm N}(x,\delta )\cap A^c\ne \varnothing$ .

Prove:

Considering $A$ as open then it will not contain any boundary point.

Suppose, $A$ will be close then its not open so it will contain all boundary points.

Alternatively, suppose $A$ will contain all boundary points then $A^c$ will not contain its boundary point.

Thus $A^c$ is open and hence $A$ is closed.

Hence, proved.