What changes could I make to these two proofs to make them stronger and or clear

Question

What changes could I make to these two proofs to make them stronger and or clearer?

1)

Here, we want to show that if x= lim a_n for some sequence (a_n) subset of A, where a_n is not equal x, then x is a limit point of A.
Because we are given that (a_n)-->x, then, by the topological definition of convergence of a sequence every V_epsilon(x) contains all but a finite number of terms of (a_n). As (a_n) subset of A, this means that V_epsilon(x) intersect A is empty set and has no elements other than x, thus x is a limit point of A

2)

Let x be an element of O, where O is an open set. If (x_n) is a sequence converging to x, prove all but a finite number of terms of (x_n) must be contained in O. (x_n)-->x. By the definiton of an open set (A set O subset of R is open if for all points a in O there exists a neighborhood V_epsilon(a) subset of O), we know that there exists V_epsilon(x) such that V_e(x) in ). Becase (x_n)-->x, by the topological definition of convergence of a sequence, every V_epsilon(x) contains all but a finite number of the terms of (x_n). Therefore, all but a finite number of the terms of (x_n) subset of O.

Explanation / Answer

At 1) perhaps because of inattention, you wrote empty instead of non-empty. See the changes in bold. Also, V_epsilon(x) intersect A has no elements other than x is not true. Its contrary is true.

1.
Because we are given that (a_n)-->x, then, by the topological definition of convergence of a sequence every V_epsilon(x) contains all but a finite number of terms of (a_n). As (a_n) subset of A and (a_n) is different than x, this means that V_epsilon(x) intersect A is non-empty set and has at least one element (in fact all but a finite) other than x, thus x is a limit point of A

For 2) I do not have corrections (except two typos). It is well done

2.

Let x be an element of O, where O is an open set. If (x_n) is a sequence converging to x, prove all but a finite number of terms of (x_n) must be contained in O. (x_n)-->x. By the definiton of an open set (A set O subset of R is open if for all points a in O there exists a neighborhood V_epsilon(a) subset of O), we know that there exists V_epsilon(x) such that V_epsilon(x) included in O ). Becase (x_n)-->x, by the topological definition of convergence of a sequence, every V_epsilon(x) contains all but a finite number of the terms of (x_n). Therefore, all but a finite number of the terms of (x_n) is a subset of O.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.