[proofplan] The equivalence constants compare the two norms pointwise, so they also compare the corresponding distances from any center $x\in V$. We first prove mutual containment of open balls, which gives equality of the two norm topologies. Then we apply the same inequalities to differences $x_n-x$ to compare sequential convergence, and to elements of a subset $S$ to compare boundedness. [/proofplan]

[step:Use the equivalence constants to compare balls around the same center]Let $m,M>0$ be constants such that, for every $v\in V$, \begin{align*} m\|v\|_a\leq \|v\|_b\leq M\|v\|_a. \end{align*} For $x\in V$ and $r>0$, define the $\|\cdot\|_a$-open ball and the $\|\cdot\|_b$-open ball by \begin{align*} B_a(x,r)=\{y\in V:\|y-x\|_a<r\} \end{align*} and \begin{align*} B_b(x,r)=\{y\in V:\|y-x\|_b<r\}. \end{align*} Fix $x\in V$ and $r>0$. If $y\in B_a(x,r/M)$, then $\|y-x\|_a<r/M$, and the upper comparison inequality gives \begin{align*} \|y-x\|_b\leq M\|y-x\|_a<r. \end{align*} Thus $y\in B_b(x,r)$, so \begin{align*} B_a(x,r/M)\subset B_b(x,r). \end{align*} If $y\in B_b(x,mr)$, then $\|y-x\|_b<mr$, and the lower comparison inequality gives \begin{align*} m\|y-x\|_a\leq \|y-x\|_b<mr. \end{align*} Since $m>0$, division by $m$ gives $\|y-x\|_a<r$, hence $y\in B_a(x,r)$. Therefore \begin{align*} B_b(x,mr)\subset B_a(x,r). \end{align*}[/step]

[guided]We want to compare the two topologies, and a norm topology is controlled by its open balls. The useful point is that the norm comparison can be applied not only to a vector $v$, but also to the difference vector $y-x$ measuring the distance from $y$ to the center $x$. Let $m,M>0$ be constants such that, for every $v\in V$, \begin{align*} m\|v\|_a\leq \|v\|_b\leq M\|v\|_a. \end{align*} For $x\in V$ and $r>0$, define \begin{align*} B_a(x,r)=\{y\in V:\|y-x\|_a<r\} \end{align*} and \begin{align*} B_b(x,r)=\{y\in V:\|y-x\|_b<r\}. \end{align*} First fix $x\in V$ and $r>0$. To put a $\|\cdot\|_a$-ball inside the $\|\cdot\|_b$-ball $B_b(x,r)$, we choose the smaller radius $r/M$. If $y\in B_a(x,r/M)$, then \begin{align*} \|y-x\|_a<r/M. \end{align*} Applying the upper inequality $\|v\|_b\leq M\|v\|_a$ to the vector $v=y-x$, we obtain \begin{align*} \|y-x\|_b\leq M\|y-x\|_a<r. \end{align*} Hence $y\in B_b(x,r)$, and therefore \begin{align*} B_a(x,r/M)\subset B_b(x,r). \end{align*} Conversely, to put a $\|\cdot\|_b$-ball inside the $\|\cdot\|_a$-ball $B_a(x,r)$, we choose the radius $mr$. If $y\in B_b(x,mr)$, then \begin{align*} \|y-x\|_b<mr. \end{align*} Applying the lower inequality $m\|v\|_a\leq \|v\|_b$ to $v=y-x$, we get \begin{align*} m\|y-x\|_a\leq \|y-x\|_b<mr. \end{align*} Since $m>0$, dividing by $m$ gives \begin{align*} \|y-x\|_a<r. \end{align*} Thus $y\in B_a(x,r)$, so \begin{align*} B_b(x,mr)\subset B_a(x,r). \end{align*}[/guided]

[step:Deduce equality of the two norm topologies from mutual ball containment] Let $\tau_a$ denote the topology induced by $\|\cdot\|_a$, and let $\tau_b$ denote the topology induced by $\|\cdot\|_b$. Let $U\in\tau_b$. For any $x\in U$, by openness in $\tau_b$ there exists $r>0$ such that \begin{align*} B_b(x,r)\subset U. \end{align*} From the ball containment proved above, \begin{align*} B_a(x,r/M)\subset B_b(x,r)\subset U. \end{align*} Thus $U$ is open in $\tau_a$, so $\tau_b\subset\tau_a$. Let $U\in\tau_a$. For any $x\in U$, by openness in $\tau_a$ there exists $r>0$ such that \begin{align*} B_a(x,r)\subset U. \end{align*} From the ball containment proved above, \begin{align*} B_b(x,mr)\subset B_a(x,r)\subset U. \end{align*} Thus $U$ is open in $\tau_b$, so $\tau_a\subset\tau_b$. Therefore $\tau_a=\tau_b$. [/step]

[step:Compare sequential convergence by applying the norm inequalities to differences] Let $(x_n)$ be a sequence in $V$, and let $x\in V$. Assume $x_n\to x$ with respect to $\|\cdot\|_a$. This means \begin{align*} \lim_{n\to\infty}\|x_n-x\|_a=0. \end{align*} For every positive integer $n$, the upper comparison inequality gives \begin{align*} 0\leq \|x_n-x\|_b\leq M\|x_n-x\|_a. \end{align*} Let $\varepsilon>0$. Since $x_n\to x$ with respect to $\|\cdot\|_a$ and $M>0$, there exists a positive integer $N$ such that $\|x_n-x\|_a<\varepsilon/M$ whenever $n\geq N$. Hence $\|x_n-x\|_b<\varepsilon$ whenever $n\geq N$, so $x_n\to x$ with respect to $\|\cdot\|_b$. Conversely, assume $x_n\to x$ with respect to $\|\cdot\|_b$. Then \begin{align*} \lim_{n\to\infty}\|x_n-x\|_b=0. \end{align*} For every positive integer $n$, the lower comparison inequality gives \begin{align*} 0\leq \|x_n-x\|_a\leq m^{-1}\|x_n-x\|_b. \end{align*} Let $\varepsilon>0$. Since $x_n\to x$ with respect to $\|\cdot\|_b$ and $m>0$, there exists a positive integer $N$ such that $\|x_n-x\|_b<m\varepsilon$ whenever $n\geq N$. Hence $\|x_n-x\|_a<\varepsilon$ whenever $n\geq N$, so $x_n\to x$ with respect to $\|\cdot\|_a$. [/step]

[step:Compare boundedness of subsets using the same constants] Let $S\subset V$. Assume $S$ is bounded with respect to $\|\cdot\|_a$. Then there exists $R_a\geq 0$ such that, for every $s\in S$, \begin{align*} \|s\|_a\leq R_a. \end{align*} Using the upper comparison inequality, for every $s\in S$, \begin{align*} \|s\|_b\leq M\|s\|_a\leq MR_a. \end{align*} Since $MR_a\geq 0$, the set $S$ is bounded with respect to $\|\cdot\|_b$. Conversely, assume $S$ is bounded with respect to $\|\cdot\|_b$. Then there exists $R_b\geq 0$ such that, for every $s\in S$, \begin{align*} \|s\|_b\leq R_b. \end{align*} Using the lower comparison inequality, for every $s\in S$, \begin{align*} \|s\|_a\leq m^{-1}\|s\|_b\leq m^{-1}R_b. \end{align*} Since $m^{-1}R_b\geq 0$, the set $S$ is bounded with respect to $\|\cdot\|_a$. We have proved equality of the two induced topologies, equivalence of sequential convergence, and equivalence of boundedness. This completes the proof. [/step]