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Epsilon Principle

Last updated Nov 1, 2022

# Statement

Let $a,b \in \mathbb{R}$ . Suppose for all $\epsilon > 0$ , we have that $a \geq b - \epsilon$ . Then $a \geq b$ .

# Proof

Suppose not. Then $a < b$ and there exists some $\delta > 0$ s.t. $a = b - \delta$ . But then $a < b - \frac{\delta}{2}$ which contradicts

our assumption that

a \geq b - \epsilon

for all

\epsilon > 0

.

\blacksquare

# Other Outlinks

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Backlinks

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