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.
I am reading Tom Apostol's Analysis and come across this theorem.   Should $a \leq b$ if $a\leq b+\epsilon$ for all $\epsilon &gt;0$?   I don't doubt the proof in the book but I don't understand the intuition or geometric explanation behind this. Could somebody shed some light on this equation? I just started studying analysis on my own.$\ \ $
It is well know that $SO(n)$ is connected and $O(n)$ has two connected components: $O^+(n)=\{A\in O(n):\det A=+1\}$ and $O^-(n)=\{A\in O(n):\det A=-1\}$.   In what book can I find this property?