拓扑学代写

If we have two elements $U \times V$ and $U^{\prime} \times V^{\prime}$ of $\mathcal{B} \Pi$, where $U$ and $U^{\prime}$ are open in $X$ and $V$ and $V^{\prime}$ are open in $Y$, then their intersection is $\left(U \cap U^{\prime}\right) \times\left(V \cap V^{\prime}\right)$, which is in $\mathcal{B}{\Pi}$ as well. Hence, for any two elements of $\mathcal{B} \Pi$ and any $a \in X \times Y$ in their intersection, there exist an element of $\mathcal{B}{\Pi}$ that contains $a$ and that is in the intersection, namely take the intersection itself. Hence, $\mathcal{B}{\Pi}$ is a basis. b. Again, to check this, checking that $\mathcal{B}{\mathrm{II}}$ is closed under finite non-empty intersections is enough. Intersections of two sets, $U \times{0}$ and $U^{\prime} \times{0}$, where $U$ and $U^{\prime}$ are open in $X$, is equal to $U \cap U^{\prime} \times{0}$, which is again in $\mathcal{B}{\mathrm{I}}$. It is similar for two sets of the type $V \times{1}$ in $\mathcal{B}{\mathrm{I}}$. On the other hand, $U \times{0} \cap V \times{1}=\emptyset, \mathcal{B}{\coprod}$ satisfies the desired property; hence, it is a basis. c. The elements of $\mathcal{B}{\Pi}$ are open in the metric topology: If $(x, y) \in U \times V$, where $U$ and $V$ are open, then there exist $\epsilon, \epsilon^{\prime}>0$ such that $B(x, \epsilon) \subset U$ and $B\left(y, \epsilon^{\prime}\right) \subset V$. Then it is easy to show that $B\left((x, y), \min \left{\epsilon, \epsilon^{\prime}\right}\right) \subset$ $U \times V$. As $(x, y) \in U \times V$ was arbitrary, this shows that the sets $U \times V$ are metric open. Thus, all the basis elements are metric open and the topology they generate is contained in $\tau_{d}$, the metric topology. On the other hand, as open subsets in metric topology are unions of ball, it is enough to show that the balls are open in the product topology. Given $(x, y)=b \in B\left(a, \epsilon^{\prime}\right) \subset \mathbb{R}^{2}$, we can find $\epsilon>0$ such that $B(b, \epsilon) \subset B\left(a, \epsilon^{\prime}\right)$. Then, clearly $(x-\epsilon / 2, x+\epsilon / 2) \times(y-\epsilon / 2, y+\epsilon / 2) \subset B((x, y), \epsilon)=B(b, \epsilon)$. As $b, a, \epsilon^{\prime}$ was arbitrary, this shows the open balls are product open, hence metric open subsets are open in the product topology. Thus, they are the same.