To prove this by contradiction, let's assume the opposite of what we want to show.

We are given that $y = \inf S$, where $S$ is a subset of real numbers. The goal is to prove that, for each $\epsilon > 0$, there exists an $a \in S$ such that $y \leq a < y + \epsilon$.

Proof by Contradiction

1. Assume the Negation: Assume, contrary to what we want to show, that there exists an $\epsilon > 0$ such that for all $a \in S$, $a \geq y + \epsilon$.

2. Implication of the Assumption: This assumption implies that $y + \epsilon$ is a lower bound of $S$, meaning all elements $a \in S$ satisfy $a \geq y + \epsilon$.

3. Contradiction with $y = \inf S$: Since $y = \inf S$, by definition, $y$ is the greatest lower bound of $S$. However, if all elements of $S$ satisfy $a \geq y + \epsilon$, then $y + \epsilon$ would be a lower bound for $S$ that is greater than $y$.

   This contradicts the assumption that $y$ is the infimum of $S$, as it implies there exists a lower bound $y + \epsilon > y$, which would make $y$ not the greatest lower bound.

4. Conclusion: Therefore, our assumption must be false. Consequently, for each $\epsilon > 0$, there must exist some $a \in S$ such that $y \leq a < y + \epsilon$.

This completes the proof by contradiction.

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Related Questions

1. How does the concept of infimum differ from minimum, and how does it apply here?
2. What other methods can be used to prove statements about infimum and supremum?
3. Why is proof by contradiction effective in cases involving bounds?
4. How does the epsilon argument generalize to other proofs involving limits or bounds?
5. What role does the assumption $\epsilon > 0$ play in this proof?

Tip:

In proofs by contradiction involving bounds, always check if the negation implies a new bound that contradicts the definition of infimum or supremum.