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

Question

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.

Answer 1

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.

Answer 2

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 $c \leq a$.