Let S be a closed set of R, T a compact set of R such that S∩ T≠Ø. ThenS∩ T isa)closed but not compactb)not closedc)compactd)neither closed nor compactCorrect answer is option 'C'. Can you explain this answer?

Question

Let S be a closed set of R, T a compact set of R such that S&#8745; T&#8800;&#216;. ThenS&#8745; T isa)closed but not compactb)not closedc)compactd)neither closed nor compactCorrect answer is option 'C'. Can you explain this answer?

Answer

Explanation:

To prove that the statement 'S ∩ T' is compact, we need to show that it is closed and bounded.

1. Closed:
Let's assume x is a limit point of 'S ∩ T'. This means that every open neighborhood of x contains a point of 'S ∩ T' other than x itself. Since x is a limit point of 'S', every open neighborhood of x contains a point of 'S'. Similarly, since x is a limit point of 'T', every open neighborhood of x contains a point of 'T'. Therefore, every open neighborhood of x contains a point of 'S ∩ T'. So, x is a limit point of 'S ∩ T', which implies that 'S ∩ T' is closed.

2. Bounded:
Since T is compact, it is bounded. Let M be a positive number such that |t| ≤ M for all t in T. Since T is a subset of R, it is also bounded in R. Let L be a positive number such that |s| ≤ L for all s in S. Therefore, every element of 'S ∩ T' is bounded by both M and L. Hence, 'S ∩ T' is bounded.

Therefore, 'S ∩ T' is closed and bounded, which implies that it is compact. Hence, the correct answer is option C - compact.

Summary:
- To prove that 'S ∩ T' is compact, we need to show that it is closed and bounded.
- We can show that 'S ∩ T' is closed by proving that every limit point of 'S ∩ T' is contained in 'S ∩ T'.
- We can show that 'S ∩ T' is bounded by using the bounds of sets S and T.
- Therefore, 'S ∩ T' is closed and bounded, which implies that it is compact.

Answer

C

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