Key Concepts and Formulas

• Negation of "for all x, P(x)" is "there exists x, such that not P(x)". Symbolically, $\sim (\forall x, P(x)) \equiv \exists x, \sim P(x)$.
• Negation of "there exists x, such that P(x)" is "for all x, not P(x)". Symbolically, $\sim (\exists x, P(x)) \equiv \forall x, \sim P(x)$.
• The negation of $x \ge M$ is $x < M$.

Step-by-Step Solution

Step 1: Identify the original statement and its components.

The given statement is: "For all $M > 0$, there exists $x \in S$ such that $x \ge M$." We can express this symbolically as: $\forall M > 0, \exists x \in S : x \ge M$.

Step 2: Negate the outer quantifier.

The original statement starts with "for all $M > 0$". The negation of "for all" is "there exists". So, the negation will start with "there exists $M > 0$". Symbolically: $\exists M > 0, \sim (\exists x \in S : x \ge M)$.

Step 3: Negate the inner quantifier.

Now we need to negate "there exists $x \in S$ such that $x \ge M$". The negation of "there exists" is "for all". So, the negation of "there exists $x \in S$ such that $x \ge M$" is "for all $x \in S$, $x < M$". Symbolically: $\exists M > 0, \forall x \in S, \sim (x \ge M)$.

Step 4: Negate the inequality.

The negation of $x \ge M$ is $x < M$. Therefore, the complete negation is "there exists $M > 0$, such that for all $x \in S$, $x < M$". Symbolically: $\exists M > 0, \forall x \in S, x < M$.

Step 5: Convert the symbolic form back to English. The statement $\exists M > 0, \forall x \in S, x < M$ translates to "there exists $M > 0$, such that $x < M$ for all $x \in S$".

Common Mistakes & Tips

• Remember to negate every part of the statement, including quantifiers and inequalities.
• The order of quantifiers is important. Changing the order can drastically change the meaning of the statement.
• Be careful with inequalities. The negation of "greater than or equal to" is "strictly less than".

Summary

We are asked to find the negation of the statement "for all M > 0, there exists x$\in$S such that x $\ge$ M". We negate the quantifiers one by one, changing "for all" to "there exists" and vice versa, and also negate the inequality. This results in the statement "there exists M > 0, such that x < M for all x$\in$S".

Final Answer

The final answer is \boxed{there exists M > 0, such that x < M for all x$\in$S}, which corresponds to option (A).