Let $A\subset \mathbb{R}$ such that $l=\text{inf }(A)$ exists. Prove that $\forall \epsilon >0 $ there is $a\in A$ in the interval $[l,l+\epsilon)$

Question

I need to prove the following:

Let $A\subset \mathbb{R}$ such that $l=\text{inf}(A)$ exists. Prove that $\forall \epsilon >0 $ there is $a\in A$ in the interval $[l,l+\epsilon)$

That's what I did:

Since $l$ is the infimum of $A$, then $l\le a, \forall a\in A$. Now, in the interval $[l, l+\epsilon)$ we can always pick the midpoint, which is $$\frac{l+l+\epsilon}{2}$$

Clearly $$l<l+\frac{\epsilon}{2}=\frac{2l+\epsilon}{2} = l+\frac{\epsilon}{2}<l+\epsilon$$

Am i right?

Answer 1

Everything you've written is correct, but you never explained how that proves the assertion - after all, why should we care about the midpoint of this particular interval? It's probably not in $A$, so it's not really relevant.

The point is that if there were no elements of $A$ in this interval, then $l + \epsilon$ is also a lower bound of $A$.

Answer 2

The easiest way to do this would be by contradiction. Suppose that given any $\epsilon>0$, there is no $a \in [l, l+ \epsilon)$. Then it must be the case that for all $a \in A, a \geq l+ \epsilon$, but this is a contradiction to the definition of $l$.