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What is the contrapositive of the following statement. Sometimes I find it confusing to form the contrapositive, so i'm interested in knowing how you formed it.

(a)$X$ is closed and bounded

(b)Given any sequence $(a_n)$ of real numbers which takes values in $X$, there exists a subsequence $(a_{n_i})$ of the original sequence, which converges to some number $L$ in $X$.

I am confused how to negate (b).

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The contrapositive of a statement $A \to B$ is $\neg B \to \neg A.$

Let $(a_n) \subset \mathbb{R}$. In your case, $A$ is the statement $(a_n) \subset X$ and $B$ the statement $\exists (a_{n_i}): \lim_{i \to \infty} a_{n_i} = L \in X$. The negation of $A$ therefore is $(a_n) \nsubseteq X$ and the negation of $B$ is $\nexists (a_{n_i}): \lim_{i \to \infty} a_{n_i} = L \in X$. Therefore, the contrapositive of statement (b) is:

If for some sequence $(a_n) \subset \mathbb{R}$ there exists no subsequence which converges to some number $L$ in $X$, then $(a_n) \nsubseteq X$.

EDIT: After having read the original question again, I think we are actually trying to negate the whole statement (b). This can be done using the common rules of logic, noting that

$$(A \to B) \iff (\neg A \vee B).$$

Using this equivalence, the negation of an implication is given by

$$\neg(A \to B) \iff \neg(\neg A \vee B) \iff A \wedge \neg B.$$

The statements $A$ and $B$ are the same as in my original answer, and the negation of $B$ is given too.

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