Suppose that $\tau_1$ and $\tau_2$ are topologies on $X$ induced by metrices $d_1$ and $d_2$, respectively, such that $\tau_1 \subseteq \tau_2$. Suppose that $\tau_1$ and $\tau_2$ are topologies on $X$
induced by metrices $d_1$ and $d_2$, respectively, such that $\tau_1 \subseteq \tau_2$. Then which of the following is TRUE?
(a) Every open ball in $(X,d_1)$ is an open ball in $(X,d_2)$.
(b) The map $x \to x$ from $(X,d_1)$ to $(X,d_2)$ is continuous.
(c) If a sequence converges in $(X,d_2)$ then it converges in $(X,d_1)$.
(d) If a sequence converges in $(X,d_1)$ then it converges in $(X,d_2)$.
Solution:
Option (d) is clearly false, since the condition of continuity holds iff $\tau_2 \subseteq \tau_1$.
I know what  we mean by convergent sequences in an arbitrary topological space. But I am unable to check the other options. I think option (a) is true but the answer key says only option (c) is true.
 A: No a) is not true. Consider the euclidean norm and the $\max$ norm, or any other $p$ norm. The norms are equivalent, so the topologies are the same. But the balls are different. Just c) is true. What you can say about a) is that for every $d_1$ ball $B_1$ there exists a $d_2$ ball included in $B_1$.
A: Every open ball in $\langle X,d_1\rangle$ is an open set in $\langle X,d_2\rangle$, but it need not be an open ball. Let $d_2$ be the discrete metric on $\Bbb R$, so that $\tau_2=\wp(\Bbb R)$, i.e., every subset of $\Bbb R$ is open:
$$d_2(x,y)=\begin{cases}
1,&\text{if }x\ne y\\
0,&\text{if }x=y\,.
\end{cases}$$
The open $d_2$-balls are the sets
$$\begin{align*}
B(x,r)&=\{y\in\Bbb R:d_2(x,y)<r\}\\
&=\begin{cases}
\{x\},&\text{if }r\le 1\\
\Bbb R,&\text{if }r>1\,.
\end{cases}
\end{align*}$$
Let $x_1$ be the usual metric on $\Bbb R$: $d_1(x,y)=|x-y|$. Then $B(0,1)$ is the open interval $(-1,1)$, which is neither a singleton $\{x\}$ nor the whole real line, so it’s not a $d_2$-ball, even though it is open in the topology $\tau_2$.
The same example shows that (b) and (d) can be false. If $f:\Bbb R\to\Bbb R:x\mapsto x$, then $\{0\}$ is open in $\langle\Bbb R,d_2\rangle$, but $f^{-1}[\{0\}]=\{0\}$, which is not open in $\langle\Bbb R,d_1\rangle$. And the sequence $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ converges to $0$ in $\langle\Bbb R,d_1\rangle$ but does not converge to anything in $\langle\Bbb R,d_2\rangle$.
