Show that $(1,1,1,...)$ is a limit point of a set $A$. 
Let $X_j$ be $\{0,1\}$, the 2 point set, with the discrete topology for $j = 1,2,…$. Let $X$ be the countable product of the $X_j$'s with the product topology. Let A be the set which consists of the following points: $(1,0,0,0,,,,),(1,1,0,0,…),...,(1,1,1,0…)$ etc. Show that $(1,1,1,1,…)$ is a limit point of A.

Define the sequence $\{a_n\}_{n \in \Bbb{N}}$ such that $a_n = (1, 1, 1, ... , 1, 0, 0, 0,...)$ where the last "$1$" is in the $n$th position. Then we know that $a_n \rightarrow (1,1,1,...)$ as $n \rightarrow \infty$. This means that $(1,1,1,...)$ is a limit of $A$ by a proposition. 
Is my answer correct? I feel like it's wrong because its too short. 
 A: Define $\textbf x = (1, 1, 1, \ldots) \in X$ and $A := \{ \textbf x_n := (1, 1, \ldots, 1, 0, 0, \ldots)$ where $\textbf x_n(i) = 0$ for $i > n\}$. 
Clearly $\textbf x \notin A$.
Claim : $\textbf x_n \to \textbf x$ in $X$. 
Let $U \ni \textbf x$ be open in $X$, and say, without loss of generality, that it is a basis element, $\displaystyle U = \prod_{i \in \mathbb N} U_i$ with $U_i \subseteq X_j$ open and $U_i = X_j$ for all but a finite number of indices. We can write 
$$
U = U_1 \times U_2 \times \cdots \times U_N \times X_{N + 1} \times X_{N + 2} \times \cdots.
$$
Observe that since $U \ni \textbf x$, it must follow that $\operatorname{pr}_i(\textbf x) = 1 \in U_i$. So our two possibilities for $U_i$ are $\{1\}$ or $X_i$. We could have a problem with say $\textbf x_1 = (1, 0, 0, \ldots)$ if $U_2 = \{1\}$. But if $n > N$, then $\textbf x_n \in U$ as we've skipped over the elements in the sequence that might have been problematic. Conclude that $\textbf x_n \to \textbf x$ in $X$. 
Hence $\textbf x$ is a limit point of $A$.

There are alternative characterizations of convergence of sequences in the product topology which you could have used if you had them at your disposal. For instance, see Munkres $\S19$ Problem $6$.
