Inequalities in definitions of convergence?

Question

Apparently in the following definitions (1) and (2) are equivalent and (3) and (4) are equivalent too!

1. $x_n$ converges to $x$ if for every $\epsilon > 0$ there's $N \in \mathbb{N}$ such that for all $n \ge N$ we have $|x_n-x|< \epsilon$.

2. $x_n$ converges to $x$ if for every $\epsilon > 0$ there's $N \in \mathbb{N}$ such that for all $n > N$ we have $|x_n-x| \le \epsilon$.

3. $x_n$ converges to $+\infty$ if for every $M > 0$ there's $N \in \mathbb{N}$ such that for all $n \ge N$ we have $X_n > M$.

4. $x_n$ converges to $+\infty$ if for every $M > 0$ there's $N \in \mathbb{N}$ such that for all $n > N$ we have $X_n \ge M$.

The same goes for convergence to $-\infty$ etc. Why does making the inequality $n \ge N$ strict i.e. saying $n \ge N$ turn the inequality $|x_n-x| < \epsilon$ into non-strict, i.e. $|x_n-x| \le \epsilon$, turning (1) to (2) and so on?

Answer 1

Regarding (1) and (2), given epsilon, you can go back and pick $N$ suitably.

Don't pick your epsilon - you seem to think, based on your comment above, that you can pick epsilon and then epsilon / 2. Let epsilon be given.