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If $\{a_n\}$ is a sequence in which $a_n \geq c$ for some constant $c$ and $a_n \rightarrow a$ then $a \geq c$

I just wanted some feedback on whether my proof of the claim is sound.

Proof

Let $\epsilon > 0$ and $ a < c$. Now $a_n \rightarrow a$ means:

$$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$$

Consider $\epsilon = \frac{c-a}{2}$

$$\Rightarrow c \leq a_n \leq a + \frac{c-a}{2} = \frac{a+c}{2} < \frac{c + c}{2} = c$$

$c < c$ is a contradiction. Therefore $a \geq c$ for the statement to hold.

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  • $\begingroup$ Your proof is good but as an aside I have to wonder about "$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$". This is maybe an editorial comment, but nobody likes to read such symbol soup an it doesn't make math more "serious". Also what you wrote doesn't parse. $|a_n-a|<\epsilon\iff a-\epsilon\le a_n\le a+\epsilon$ is always true for all $n$, $a_n$, $a$ and $\epsilon$. Does that mean all sequences converge to all values? $\endgroup$ – fleablood Oct 21 '18 at 15:14
  • $\begingroup$ Funny you mention that. I was talking to my professor about this same thing this week because I'm not a big fan of the "symbol soup" as you call it and was inquiring whether just communicating my ideas in plain English would be better. $\endgroup$ – dc3rd Oct 21 '18 at 15:18
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The idea is fine but:

  • You should not begin with “Let $\varepsilon>0$”. You fix $\varepsilon$ later, not at this point.
  • You should write that you are assuming that $a<c$, in order to get a contradiction. You can't just say “Let […] $a<c$”, because $a$ and $c$ are fixed from the start.
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  • $\begingroup$ Should I state it in terms of: "assume $a < c$ towards a contradiction....."? And with regards to $\epsilon$ don't I have to declare at the start that it will always be positive? $\endgroup$ – dc3rd Oct 21 '18 at 15:02
  • $\begingroup$ This is a matter of taste, but I would begin with “In order to get a contradiction, let us assume that $a<c$”. Concerning the other question, note that you wrote right after $\forall\varepsilon>0$. So, no, you don't have to write that. $\endgroup$ – José Carlos Santos Oct 21 '18 at 15:04
  • $\begingroup$ Gracias por la ayuda. $\endgroup$ – dc3rd Oct 21 '18 at 15:07
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Your proof seems legitimate to me, but it is a proof by contradiction, so here is an alternate way to do it directly. I will start with what you have.

$$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$$

Now, this implies that for any $\epsilon > 0$, $a_n \leq a+\epsilon$ for some $n \in \Bbb{N}$. Also, $c \leq a_n$, so we get $c\leq a+\epsilon$ for any $\epsilon > 0$. Thus, $c\leq g$ for all real numbers $g$ in the interval $(a, \infty)$. It is well known that $a$ is the greatest lower bound for $(a, \infty)$, so this implies $c \leq a$.

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