Question: Show that if $a \leq x_n \leq b$ for every $n$ and $x_n \rightarrow x$, then $a \leq x \leq b$.
Proof: Let $\epsilon>0$. By assumption $a_n \leq x_n \leq b$ for all $n$. By definition of convergence, we have that there exists an $N_1$ such that $|x_n-x|<\epsilon $ for $n \geq N_1$.
Thus, when $n \geq N_1$, we have that $$x-\epsilon < x_n \leq b$$ and $$a \leq x_n < x+ \epsilon$$ Since $\epsilon$ is arbitrary, we have: $$x < b$$ and $$a<x$$ Thus, $a<x<b$. Is my proof close to being correct, or am I not on the right track. Some confusions I'm having.
- Can I say that $\epsilon$ being arbitrary implies that $x<b$ and $a<x$.? Or is this an inaccurate statement? I have seen this argument in other proofs but don't fully understand the implication.
- I proved that $a<x<b$ but not $a \leq x \leq b$. How do I prove it for $\leq$ and $\geq$? Thanks and sorry for the seemingly trivial and basic questions.