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.

  1. 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.
  2. 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.

Note that from $|x_n-x|<\epsilon$ you can write $$x<x_n+\epsilon\le b+\epsilon\\a-\epsilon\le x_n-\epsilon<x$$therefore $$a-\epsilon<x<b+\epsilon$$Since this holds for any $\epsilon >0$, we obtain $$a\le x\le b$$ This is where the $\le$ kicks in.

  • $\begingroup$ Hi @Mostafa Ayaz isn't this just the same exact thing I showed? I don't see the difference, could you let me know the difference and where I messed up. But doesn't $a- \epsilon< x<b+\epsilon$ holding for any $\epsilon >0$ imply $a<x<b$? i don't see how it implies $a \leq x \leq b$. thank you $\endgroup$ – kemb Aug 24 '19 at 22:39
  • $\begingroup$ Yes. This is almost exactly what you proved, except the last conclusion. For example let $$1-{1\over n}<x<2+{1\over n}$$ hold for any positive integer $n$. Does this mean $$x\in[1,2]$$ or $x\in (1,2)$? i.e. does $x=1$ or $x=2$ satisfy the inequality? $\endgroup$ – Mostafa Ayaz Aug 24 '19 at 22:49
  • $\begingroup$ Hi @Mostafa Ayaz. thats what I don't understand. I don't understand why $1- \frac{1}{n}<x< 2+\frac{1}{n}$ holding for any positive integer n implies that $x \in [1,2]$. I thought it implied that $x \in (1,2)$. Could you explain why it implies that $x \in [1,2]$ as I don't understand this concept. Sorry for the basic questions $\endgroup$ – kemb Aug 24 '19 at 22:56
  • $\begingroup$ It is OK. Do you agree that any $x_0$ for which the inequality holds for $n\in \Bbb N$, falls in the limit interval? $\endgroup$ – Mostafa Ayaz Aug 24 '19 at 22:57
  • $\begingroup$ Yes I agree with that @Mostafa Ayaz $\endgroup$ – kemb Aug 24 '19 at 22:58

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