(a) Since $\mathbb{R}$ is path connected, $\mathbb{R}^\omega$ in the product topology is path connected and connected (§24, exercise 8(a)), therefore, there is one component = one path component, i.e. the whole space.(b) $f(\mathbf{x})=\mathbf{x}-\mathbf{x_0}$ preserves the metric, and therefore, by exercise 2 of §21, it is a homeomorphism. Hence, $\mathbf{x}$ and $\mathbf{y}$ are in the same component iff $\mathbf{z}=\mathbf{x}-\mathbf{y}$ and $\mathbf{0}=(0,0, . . .)$ are in the same component. We show that it is iff $\mathbf{z}$ is bounded.First, we show that $\mathbf{0}$ is path connected to any other point $\mathbf{x}\in B_{\bar{\rho}}(\mathbf{0},\epsilon)$ where $\epsilon\le 1$ . Indeed, $f(t)_{t\in[0,1]}=t\mathbf{x}$ is a path. To see this, recall that a function to the subspace is continuous iff it is continuous as a function in to the space, and that the topology in a subspace is generated by the same metric as the topology of the space. Moreover, $t'\mathbf{x}\in B_{\bar{\rho}}(t\mathbf{x},r)$ iff $sup_i\{|t'-t|x_i\}<r$ iff $|t'-t|<r/sup_i\{x_i\}$ . Therefore, the preimage is open.Now, two things follow. First, the space is locally path connected. Therefore, two points are connected iff they are path connected. Second, $\mathbf{0}$ and $\mathbf{x}$ are path connected iff $\mathbf{x}$ is bounded. Indeed, according to the exercise 8 of §23, the sets of bounded and unbounded sequences is a separation, therefore, any point connected to $\mathbf{0}$ (which is bounded) must be bounded. The other direction: if a sequence is bounded, then the argument similar to one in the previous paragraph shows that there is a path.(c) If $\mathbf{x}-\mathbf{y}\in\mathbb{R}^\infty$ then $f(t)_{t\in[0,1]}=\mathbf{x}+t(\mathbf{y}-\mathbf{x})$ is a path from $\mathbf{x}$ to $\mathbf{y}$ . Indeed, there is a finite number of nonzero coordinates in $\mathbf{y}-\mathbf{x}$ , and, therefore, the preimage of any open basis set of the box topology is a finite intersection of intervals open in $[0,1]$ . On the other hand, if $\mathbf{x}-\mathbf{y}\notin\mathbb{R}^\infty$ then there is an infinite sequence of indexes $j$ such that $x_j\neq y_j$ and the homeomorphism $h(\mathbf{z})_i=z_i-x_i$ if $x_i=y_i$ or $h(\mathbf{z})_i=n(z_i-x_i)/(y_i-x_i)$ otherwise (it is a homeomorphism by exercise 8 of §19), maps $\mathbf{x}$ to $\mathbf{0}$ and $y$ to an unbounded sequence. But the sets of bounded and unbounded sequences are open in the box topology (they are open even in the coarser uniform topology) and form a separation. Therefore, $\mathbf{x}$ and $\mathbf{y}$ must be in different components. BTW, the result also implies that $\mathbb{R}^\omega$ in the box topology is not locally path connected, still components and path components are the same.