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

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.

• 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? – fleablood Oct 21 '18 at 15:14
• 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. – dc3rd Oct 21 '18 at 15:18

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 $$a, in order to get a contradiction. You can't just say “Let […] $$a”, because $$a$$ and $$c$$ are fixed from the start.
• 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? – dc3rd Oct 21 '18 at 15:02
• 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. – José Carlos Santos Oct 21 '18 at 15:04
• Gracias por la ayuda. – dc3rd Oct 21 '18 at 15:07

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 $$c \leq a$$.