# Notes - Analysis I MT22, Approximation property

### Flashcards

When proving the ‘$\impliedby$‘ direction of the approximation property:

Let $A\subseteq\mathbb{R}$. Prove $\sup(A)=\alpha$ if and only if:

- $\alpha$ is an upper bound of $A$, and
- Given any $\epsilon > 0$, there exists some $x \in A$ such that $\alpha - \epsilon < x$.

What contradiction do you use?

Suppose $\alpha$ and $\beta$ are both upper bounds of $A$. Assume $\beta < \alpha$.

When trying to prove that if $\alpha = \sup A$ then given any $\varepsilon > 0$, there exists $x \in A$ such that $\alpha - \varepsilon < x$, what do you consider?

Whether $\alpha - \varepsilon$ could be an upper bound.