Proof explanation: If $a_n \leq c \text{ } \forall \text{ } n \in \mathbb{N}$ for $c \in \mathbb{R}$, then $\lim_{n\to\infty} a_n \leq c$

Question

I have difficulties understanding this proof.

Let $(a_n)_{n \in \mathbb{N}}$ be a convergent sequence with $\lim_{n\to\infty} a_n = a$.

Prove: If $a_n \leq c \text{ } \forall \text{ } n \in \mathbb{N}$ for $c \in \mathbb{R}$, then $a \leq c$.

Proof:

Suppose $a > c \Rightarrow a-c > 0$

Since $a_n \to a$ there exists $\forall \epsilon > 0$ an $N \in \mathbb{N}$ with $|a_n-a| < \epsilon \text{ } \forall n > N,$ especially for $\epsilon = a-c$

$$\Rightarrow |a_n-a| < a-c \text{ } \forall n > N \\ \Leftrightarrow - (a-c) < a_n -a < a-c \text{ } \forall n > N \\\Leftrightarrow c<a_n \forall n \in \mathbb{N}$$

I already have troubles understanding what is being asked for. For example, if we have $a_n = \frac{1}{n}$ and $c = 1$, is the statement implying that $\lim_{n \to \infty} a_n = 0 \Rightarrow 0 < 1$?

As for the proof, I cannot visualize in my head what is meant with $|a_n -a| < a -c$ and how we get to $c < a_n$.

Answer 1

I thin you have a typo when you say: Suppose $a<c\implies a-c=0$. The proof is by reductio ad absurdum, you assume that the result is false, that's it, $a>c$, so $a-c>0$ and you can take positive $\epsilon=a-c$ . With this assumption you end up having that for some $N\in\mathbb{N}$(that depends on the choosen $\epsilon$) we have that $a_n>c$ for $n>N$, but that contradicts our hypothesis, hence $a<c$

Answer 2

Beware: $a>c$ implies $a-c>0$, not $c-a>0$ (you have a typo, easily fixable).

How do you show that $a\le c$, where $a$ and $c$ are arbitrary real numbers? You can prove that, for every $\varepsilon>0$, it holds that $a<c+\varepsilon$.

So, take $\varepsilon>0$. Then, by definition of convergence, there exists $N$ such that, for $n>N$, it holds that $$ |a_n-a|<\varepsilon $$ that implies $$ a<a_n+\varepsilon $$ Choose $n>N$. Then $$ a<a_n+\varepsilon\le c+\varepsilon $$ because, by assumption, $a_n\le c$. Therefore we have proved that $a<c+\varepsilon$ and we're done.

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Your argument is good as well, but with some fixes. You find $N$ such that, for $n>N$, it holds that $$ |a_n-a|<a-c $$ This is equivalent to $$ c-a<a_n-a<a-c $$ and the left inequality is the same as $a_n>c$, which is false for every $n$, but should hold for all $n>N$. Contradiction.

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Yes, using this result for $a_n=1/n$, you prove that $0\le 1$. That's true, isn't it?