What is this question about convergence asking? Loomis and Sternberg, Advanced Calculus, p.369, section 9.2, exercise 2.2:

Show that a sequence $\{x_\alpha\}$ converges to $x$ if and only if for every open set $U$ containing $x$ there is an $N_U$ with $x_\alpha\in U$ for $\alpha>N_U$.

I am a bit confused about what it is asking. If it is talking about convergence of a sequence points on a topological space, this just follows directly from the definition. However, the context of this section is about convergence of a sequence in a set equipped with an atlas. Atlas is defined on p.364 as follows:

Let $M$ be a set. Let $V$ be a Banach space. ... A $V$-atlas of class $C^k$ on $M$ is a collection $\mathcal A$ of pairs $(U_i,\varphi_i)$ called charts, where $U_i$ is a subset of $M$ and $\varphi_i$ is a bijective map of $U_i$ onto an open subset of $V$ subject to the following conditions:
A1. For any $(U_i,\varphi_i)\in\mathcal A$ and $(U_j,\varphi_j)\in\mathcal A$ the sets $\varphi_i(U_i\cap U_j)$ and $\varphi_j(U_i\cap U_j)$ are open subsets of $V$, and the maps $$\varphi_i\circ\varphi_j^{-1}:\varphi_j(U_i\cap U_j)\to\varphi_i(U_i\cap U_j)$$ are differentiable of class $C^k$.
A2. $\cup U_i=M$.

And convergence of a sequence in a set equipped with an atlas is defined on p.368 as follows:

Let $M$ be a set with an atlas $\mathcal A$. We shall say that a sequence of points $\{x_i\in M\}$ converges to $x\in M$ if

*

*there exists a chart $(U_i,\varphi_i)\in\mathcal A$ and an integer $N_i$ such that $x\in U_i$ and for all $k>N_i$, $x_k\in U_i$;

*$\{\varphi_i(x_k)\}_{k>N_i}$ converges to $\varphi_i(x)$.


If the exercise is about convergence in the above sense (I don't know how to call it; is it convergence on an atlas or convergence on a manifold?), I don't know how the statement in question can possibly be proved. Note that $M$ is just a set. It is not a topological space and I don't know what the "open set" in the problem statement refers to. Also, while $\varphi_i\circ\varphi_j^{-1}$ is assumed to be of class $C^k$, $\varphi_i$ itself is not even assumed to be continuous in the definition. And while $\varphi_i(U_i\cap U_j)$ is open, the definition of atlas says nothing about $\varphi_i(U_i\cap U)$ for an arbitrary open set $U$. Any idea?
 A: I thank user "peek-a-boo" for his/her help. Right before the exercise begins, the authors have actually defined open sets on $M$. A set $U\subset M$ is defined to be open if $\varphi_i(U_i\cup U)$ is open in the Banach space $V$ for every chart $(U_i,\varphi_i)$.
With this definition, we can prove the problem statement as follows. Suppose $\{x_\alpha\}$ converges to $x$ on $M$. Let $(U_i,\varphi_i)$ be a chart that satisfies definition of convergence on $M$. Let $U\ni x$ be an open set in $M$. By definition, $\varphi_i(U_i\cap U)\ni\varphi_i(x)$ is open in $V$ and $\{\varphi_i(x_\alpha)\}$ converges to $\varphi_i(x)$. Hence there exists an integer $N$ such that $\{\varphi_i(x_\alpha)\in\varphi_i(U_i\cap U)$ for all $\alpha>N$. Therefore $x_\alpha\in U_i\cap U\subseteq U$ for all $\alpha>N$.
Conversely, suppose that for every open set $U\ni x$, there exists an integer $N$ such that $x_\alpha\in U$ for all $\alpha>N$. In particular, this is true when $U=U_i$ for any chart $(U_i,\varphi_i)$ such that $U_i\ni x$ (such a chart exists because $M=\cup_iU_i$; also, $U_i$ is open by definition). Hence condition 1 of convergence on $M$ is satisfied for some integer $N_i$. Now let $W$ be any open set in $V$ that contains $\varphi_i(x)$. Put $U=U_i\cap\varphi_i^{-1}(W)$. Then $U\ni x$ is open in $M$. By assumption, there exists an integer $N\ge N_i$ such that $x_\alpha\in U$ for all $\alpha>N$. Hence $\varphi_i(x_\alpha)\in\varphi_i(U)\subseteq W$ for all $\alpha>N$. Since $W$ is an arbitrary open set containing $\varphi_i(x)$, we conclude that $\{\varphi_i(x_\alpha)\}$ converges to $\varphi_i(x)$.
