Let S be an infinite subset of R such that S\{a} is compact for some S. Then which one of the following is TRUE ?a)S is a connected setb)S contains no limit pointsc)S is a union of open intervalsd)Every sequence in S has a subsequence converging to an element in SCorrect answer is option 'D'. Can you explain this answer?

Question

Let S be an infinite subset of R such that S\{a} is compact for some   S. Then which one of the following is TRUE ?a)S is a connected setb)S contains no limit pointsc)S is a union of open intervalsd)Every sequence in S has a subsequence converging to an element in SCorrect answer is option 'D'. Can you explain this answer?

Answer

Explanation:

Let's go through each option to determine which one is true.

a) S is a connected set:

This statement is not necessarily true. It is possible for S to be disconnected. For example, consider the set S = (0,1) U (2,3) in R. S\{2} = (0,1) U (2,2.5) U (2.5,3) is compact, but S is not connected.

b) S contains no limit points:

This statement is also not necessarily true. S can contain limit points. For example, consider the set S = [0,1] U {2} in R. S\{2} = [0,1] is compact, but S contains the limit point 1.

c) S is a union of open intervals:

This statement is not necessarily true. S can be a union of closed intervals or a combination of closed and open intervals. For example, consider the set S = [0,1] U (2,3) in R. S\{2} = [0,1] U (2,3) is compact, but S is not a union of open intervals.

d) Every sequence in S has a subsequence converging to an element in S:

This statement is true. Let's prove it.

Since S\{a} is compact, every sequence in S\{a} has a convergent subsequence.

Now, consider any sequence (an) in S.

If (an) is a constant sequence, then the subsequence (an) itself converges to an.

If (an) is not a constant sequence, then there exists a subsequence (an_k) in S\{a} that converges to some element b in S\{a}.

Since S\{a} is compact, b must also belong to S\{a}.

Therefore, the subsequence (an_k) converges to b in S.

Hence, every sequence in S has a subsequence converging to an element in S.

Therefore, the correct answer is option 'd'.

Answer

The Correct Answer is Option D: Every sequence in S has a subsequence converging to an element in S.

To understand why this is true, let's break down the options and eliminate them one by one:

a) S is a connected set:
This statement is not necessarily true for all infinite subsets S of R. There exist infinite subsets of R that are disconnected, such as the set of rational numbers. So, option a) is not always true.

b) S contains no limit points:
This statement is also not necessarily true for all infinite subsets S of R. Consider the set S = {1, 1/2, 1/3, 1/4, ...}. This set contains the limit point 0, as the sequence of elements in S converges to 0. Therefore, option b) is not always true.

c) S is a union of open intervals:
This statement is not necessarily true for all infinite subsets S of R. There exist infinite subsets of R that are not a union of open intervals, such as the set of irrational numbers. So, option c) is not always true.

Therefore, the only option remaining is:

d) Every sequence in S has a subsequence converging to an element in S:

To prove the correctness of this statement, let's assume that S\{a} is compact for some a in S. This means that every sequence in S\{a} has a subsequence converging to an element in S\{a}.

Now, consider any arbitrary sequence (xn) in S. Since S\{a} is compact, the subsequence (xn) of (xn) must have a subsequence (xnk) that converges to some element b in S\{a}. Since b is in S\{a}, it is also in S.

Therefore, every sequence in S has a subsequence (xnk) that converges to an element b in S, which proves that option d) is true.

In conclusion, option d) is the correct answer because it holds true for any infinite subset S of R for which S\{a} is compact for some a in S.

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