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So I think this is the last problem I have and I'm not thinking I'm doing it properly.
Let $a,b$ be real numbers and suppose for all $\varepsilon \gt 0, a \le b+\varepsilon$. Show that $a \le b$.
Assume $b\lt a$. Then $b+\varepsilon \lt a + \varepsilon$ and $$a \le b+\varepsilon \lt a+\varepsilon$$ If I can show that there is a contradiction I can prove it, but I'm having trouble with the contradiction.
Since you like a proof assuming $b<a$, let's do that. Consider $b+\frac{a-b}2$, halfway between $b$ and $a$.
You could prove the contrapositive of this question:
Suppose that $b<a$. Then there exists $\varepsilon>0$ so that $b+\varepsilon<a$.
Since $a\le b+\varepsilon$ for all $\varepsilon$. Then either $a\le b$ or $a>b$. Then we will show that the latter leads a contradiction. For sake of contradiction suppose that $a>b$, then $a-b>0$, and since it holds for all $\varepsilon>0$, then in particular it holds for $0<\varepsilon<a-b$. So, $a\le b+\varepsilon<b+(a-b)=a$, i.e., $a<a$. Contradiction. Thus $a\le b$ as desired.
Now try to show that if $|a-L|\le \varepsilon$ for all $\varepsilon>0$, then $a=L$
Assume that $a>b$. Then we have $a>b+\epsilon$ for $\epsilon=\frac {b-1} \epsilon$
