It may be a silly question. But I really need help to check whether my proof is correct.
It is known that in a Cartesian product norm space $X \times Y$, a sequence $(a_n, b_n)$ in $X \times Y$ converges to $(a,b)$ in $X \times Y$ if and only if $\{a_n\}$ converges to $a$ in $X$ and $\{b_n\}$ converges to $b$ in $Y$ as $n$ tends to infinity. Is it also true in topological space?
my proof Let $(a_n, b_n)$ be a sequence in $X \times Y$ converges to $(a,b)$ in $X \times Y$. Let $V_a$ be any open neighbourhood of $a$ and $V_b$ any open neighbourhood of $b$. There is a $N$ such that $(a_n, b_n) \in V_a \times V_b$ for all $n \geq N$. Thus, $(a_n) \in V_a$ and $(b_n) \in V_b$ for all $n \geq N$. It shows that $(a_n)$ converges to $a$ and $(b_n)$ converges to $b$.
Conversely, for every open neighbourhood $U_{a, b}$, there exist open sets $V_a$ in $X$ and $V_b$ in $Y$ so that $a, b \in V_a \times V_b \subseteq U_{a, b}$. Since $(a_n)$ converges to $a$ and $(b_n)$ converges to $b$. There are $N_a$, $N_b$ so that $a_n \in V_a$ for all $n \geq N_a$ and $b_n \in V_b$ for all $n \geq N_b$. Take $N = \max \{N_a, N_b\}$, then we have $(a_n, b_n) \in V_a \times V_b \subseteq U_{a, b}$ for all $n \geq N$. So, $(a_n, b_n)$ converges to $(a, b)$.