Understanding convergence of sequences in metric spaces

Question

Rudin told me that a sequence $\{p_n\}$ converges to $p$ in a metric space $X$, if for all $\epsilon > 0$ there exists $N\in \mathbb{Z}$ such that

$$n \geq N \implies d(p,p_n) < \epsilon$$

I don't understand how this corresponds to my naive "sloppy-calculus" understanding of converging sequences in $\mathbb{R}$; which if they converge, they get closer and closer to $p$ and "reaching it at infinity".

In particular I don't understand how the existence of $N$ such that $n \geq N$ could imply that the distance between $p$ and $p_n$ is less than $\epsilon$. I can't really wrap my head around this concept in the rigorous formulation.

Can someone explain how one should think about this, perhaps together with a simple example or two?

I hope this question isn't too fuzzy, I'm taking a course in real analysis and to be honest I'm really struggling with understanding the material.

Answer 1

What the statement says is: $$\forall \varepsilon > 0,\ \exists N_\varepsilon\text{ s.t. } \forall n\geq N_\varepsilon, \text{ we have } d(p_n,p) < \varepsilon $$ Let us break it down:

• $\forall \varepsilon > 0$:

no matter how close you want it to be

• $\exists N_\varepsilon$

there exists an index of the sequence (which can depend on the "how close you want" parameter $\varepsilon$)

• such that $\forall n\geq N_\varepsilon$

such that after this index, the terms of the sequence

• $d(p_n,p) < \varepsilon$

stay as close as you wanted from the limit.

The use of that index $N_\varepsilon$ is to say that sure, the first few terms of the sequence (the first 10, 45, or maybe $10^6/\varepsilon$) can be as far as $p$ as they want; but after that, they have to remain very close to the limit $p$. I.e., this definition captures the following intuitive idea:

"if you are a sequence that converges to $p$, then after a while your terms all better be close to $p$."