If A is an open set in Rn and C = {x1,x2,...,xn} is a finite set of points of A, then

[Tom202412]

If A is an open set in Rⁿ and C = {x1,x2,...,x_n} is a finite set of points of A, then:

show that:

A-B = {x E A : X (does not belong to) C} is also an open set

 

[stapel]

If A is an open set in Rⁿ and C = {x1,x2,...,x_n} is a finite set of points of A, then show that

A-B = {x E A : X (does not belong to) C} is also an open set

What is the relationship between "x" and "X"? Would it be correct to assume that "E" means "is an element of"?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!

 

[Tom202412]

What is the relationship between "x" and "X"? Would it be correct to assume that "E" means "is an element of"?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!

Pardon my ambiguity. I rephrased the statement:

If A is an open set in Rⁿ and C = {x1,x2,...,x_n} is a finite set of points of A, then:

show that:

A-B = {x ɛ A : x ∉ C} is also an open set

 

[pka]

If A is an open set in Rⁿ and C = {x1,x2,...,x_n} is a finite set of points of A, then:
show that: A-B = {x E A : X (does not belong to) C} is also an open set

You have got to mean C and not B above. Do you know that \(\displaystyle A\setminus C=A\cap C^c ~?\)
If \(\displaystyle x]in A\setminus C\) then let \(\displaystyle \delta=\min\{d(x,x_k):k=1,2,\cdots,x_n\}\). Why is \(\displaystyle \delta>0~?\)
There is a open set \(\displaystyle \mathcal{O}\) such that \(\displaystyle x\in\mathcal{O}\subset A \). Why is that?
Let \(\displaystyle \mathcal{Q}=\mathcal{O}\cap\mathcal{B}(x;\delta) \)(the ball centered at \(\displaystyle x\) with radius \(\displaystyle \delta\).

Now you give an argument that \(\displaystyle x\in\mathcal{Q}\subset A\setminus C. \)

If you want any help in the future, the you must reply with answers and or questions.