[proofplan] We compare neighborhoods at an arbitrary point $(x,y) \in X \times Y$. First, every $d_p$-ball contains a product of coordinate balls, so every $d_p$-[open set](/page/Open%20Set) is open in the [product topology](/page/Product%20Topology). Second, every basic product neighborhood contains a sufficiently small $d_p$-ball, using the coordinate bounds $d_X(x,x') \le d_p((x,y),(x',y'))$ and $d_Y(y,y') \le d_p((x,y),(x',y'))$. These two local containment statements prove equality of the two topologies. [/proofplan]

[step:Fix the two topologies and the notation for balls] Let $\tau_X$ be the metric topology on $X$ induced by $d_X$, let $\tau_Y$ be the metric topology on $Y$ induced by $d_Y$, and let $\tau_{\mathrm{prod}}$ be the product topology on $X \times Y$ generated by sets of the form $U \times V$ with $U \in \tau_X$ and $V \in \tau_Y$. Let $\tau_p$ denote the topology on $X \times Y$ induced by the [product metric](/page/Product%20Metric) $d_p$ specified in the theorem statement. For $x \in X$ and $r>0$, define the $d_X$-ball \begin{align*} B_X(x,r)=\{x' \in X : d_X(x,x')<r\}. \end{align*} For $y \in Y$ and $r>0$, define the $d_Y$-ball \begin{align*} B_Y(y,r)=\{y' \in Y : d_Y(y,y')<r\}. \end{align*} For $(x,y) \in X \times Y$ and $r>0$, define the $d_p$-ball \begin{align*} B_p((x,y),r)=\{(x',y') \in X \times Y : d_p((x,y),(x',y'))<r\}. \end{align*} [/step]

[step:Place a product of coordinate balls inside each $d_p$-ball]Fix $(x,y) \in X \times Y$ and $\varepsilon>0$. We claim that \begin{align*} B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2) \subset B_p((x,y),\varepsilon). \end{align*} Let $(x',y') \in B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2)$. Then \begin{align*} d_X(x,x')<\varepsilon/2 \end{align*} and \begin{align*} d_Y(y,y')<\varepsilon/2. \end{align*} If $1 \le p < \infty$, then \begin{align*} d_p((x,y),(x',y'))=\left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}. \end{align*} Using the two coordinate bounds gives \begin{align*} d_p((x,y),(x',y'))<\left((\varepsilon/2)^p+(\varepsilon/2)^p\right)^{1/p}. \end{align*} The right-hand side equals $2^{1/p}\varepsilon/2$. Since $p \ge 1$, we have $2^{1/p} \le 2$, and therefore \begin{align*} d_p((x,y),(x',y'))<\varepsilon. \end{align*} If $p=\infty$, then \begin{align*} d_\infty((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}<\varepsilon/2<\varepsilon. \end{align*} Thus the claimed containment holds for every $1 \le p \le \infty$. It follows that every $\tau_p$-open set is $\tau_{\mathrm{prod}}$-open: if $W \in \tau_p$ and $(x,y) \in W$, choose $\varepsilon>0$ with $B_p((x,y),\varepsilon)\subset W$. The product neighborhood $B_X(x,\varepsilon/2)\times B_Y(y,\varepsilon/2)$ lies in $W$ by the containment above, so $W$ is open in the product topology.[/step]

[guided]Fix a point $(x,y) \in X \times Y$ and a radius $\varepsilon>0$. Recall that \begin{align*} B_X(x,r)=\{x' \in X : d_X(x,x')<r\} \end{align*} denotes the open ball in $(X,d_X)$, that \begin{align*} B_Y(y,r)=\{y' \in Y : d_Y(y,y')<r\} \end{align*} denotes the open ball in $(Y,d_Y)$, and that \begin{align*} B_p((x,y),r)=\{(x',y') \in X \times Y : d_p((x,y),(x',y'))<r\} \end{align*} denotes the open ball in the product metric. To show that the $d_p$-topology is no finer than the product topology, we must show that a $d_p$-ball contains a basic product neighborhood of its center. The natural candidate is a product of two small coordinate balls: \begin{align*} B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2). \end{align*} Take an arbitrary point $(x',y')$ in this product. By membership in the first coordinate ball, \begin{align*} d_X(x,x')<\varepsilon/2. \end{align*} By membership in the second coordinate ball, \begin{align*} d_Y(y,y')<\varepsilon/2. \end{align*} Now we verify that $(x',y')$ is inside the $d_p$-ball of radius $\varepsilon$. If $1 \le p < \infty$, the definition of the product metric gives \begin{align*} d_p((x,y),(x',y'))=\left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}. \end{align*} Substituting the two strict coordinate inequalities gives \begin{align*} d_p((x,y),(x',y'))<\left((\varepsilon/2)^p+(\varepsilon/2)^p\right)^{1/p}. \end{align*} The expression on the right simplifies to $2^{1/p}\varepsilon/2$. Since $p \ge 1$, the inequality $2^{1/p}\le 2$ holds, so \begin{align*} 2^{1/p}\varepsilon/2 \le \varepsilon. \end{align*} Hence \begin{align*} d_p((x,y),(x',y'))<\varepsilon. \end{align*} For the endpoint $p=\infty$, the metric is the maximum of the coordinate distances. The same coordinate bounds give \begin{align*} d_\infty((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}<\varepsilon/2<\varepsilon. \end{align*} Therefore \begin{align*} B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2) \subset B_p((x,y),\varepsilon) \end{align*} for every $1 \le p \le \infty$. This containment converts metric openness into product openness. Let $W$ be open in $\tau_p$, and let $(x,y)\in W$. By definition of the metric topology, there exists $\varepsilon>0$ such that \begin{align*} B_p((x,y),\varepsilon)\subset W. \end{align*} The containment just proved supplies a basic product neighborhood \begin{align*} B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2) \end{align*} with \begin{align*} (x,y)\in B_X(x,\varepsilon/2) \times B_Y(y,\varepsilon/2)\subset W. \end{align*} Thus every point of $W$ has a basic product neighborhood contained in $W$, which is precisely the criterion that $W$ be open in the product topology.[/guided]

[step:Place a $d_p$-ball inside each basic product neighborhood] Let $U \in \tau_X$ and $V \in \tau_Y$, and fix $(x,y) \in U \times V$. Since $U$ is open in the metric topology induced by $d_X$, there exists $r_X>0$ such that \begin{align*} B_X(x,r_X)\subset U. \end{align*} Since $V$ is open in the metric topology induced by $d_Y$, there exists $r_Y>0$ such that \begin{align*} B_Y(y,r_Y)\subset V. \end{align*} Define \begin{align*} r=\min\{r_X,r_Y\}. \end{align*} Then $r>0$. We claim that \begin{align*} B_p((x,y),r)\subset U \times V. \end{align*} Let $(x',y') \in B_p((x,y),r)$. If $1 \le p < \infty$, then \begin{align*} d_X(x,x') \le \left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}=d_p((x,y),(x',y'))<r. \end{align*} Also, \begin{align*} d_Y(y,y') \le \left(d_X(x,x')^p+d_Y(y,y')^p\right)^{1/p}=d_p((x,y),(x',y'))<r. \end{align*} If $p=\infty$, then \begin{align*} d_X(x,x')\le d_\infty((x,y),(x',y'))<r \end{align*} and \begin{align*} d_Y(y,y')\le d_\infty((x,y),(x',y'))<r. \end{align*} In all cases, $d_X(x,x')<r\le r_X$ and $d_Y(y,y')<r\le r_Y$, so $x'\in U$ and $y'\in V$. Hence $(x',y')\in U\times V$. Therefore every basic product neighborhood contains a $d_p$-ball around each of its points. Since arbitrary product-open sets are unions of such basic product neighborhoods, every $\tau_{\mathrm{prod}}$-open set is $\tau_p$-open. [/step]

[step:Conclude that the two topologies are equal] The previous two steps prove the two inclusions \begin{align*} \tau_p \subset \tau_{\mathrm{prod}} \end{align*} and \begin{align*} \tau_{\mathrm{prod}} \subset \tau_p. \end{align*} Therefore \begin{align*} \tau_p=\tau_{\mathrm{prod}}. \end{align*} Thus the topology induced by $d_p$ on $X\times Y$ is exactly the product topology induced by the metric topologies on $X$ and $Y$. [/step]