Take (b_{1}, b_{2}, ...) ∈ R^{ω} such that there exists an M such that |b_{i}| ≤ M for all i. Define f : [0, 1] → R^{ω} by f(t) = (b_{1}t, b_{2}t, ...). If f is continuous then f is a path from (0, 0, ... ) to (b_{1}, b_{2}, ...). Take V open in R^{ω} and take x ∈ f^{−1}(V). V open, f(x) ∈ V implies there exists an r > 0 such that B\overline{ρ}(f(x), r) ⊂ V. f(x) = (b_{1}x, b_{2}x, ....). Claim: (x − \frac{r}{2M} , x + \frac{r}{2M}) ⊂ f^{−1}(B\overline{ρ}(f(x), r) Take s ∈ (x − \frac{r}{2M} , x + \frac{r}{2M}). f(s) = (b_{1}s, b_{2}s, ...) |s − x| < \frac{r}{2M} implies |b_{i}s − b_{i}x| < \frac{r|b_{i}|}{2M} ≤ \frac{r}{2} Thus sup|b_{i}s − b_{i}x| ≤ \frac{r}{2} < r. Hence f(s) ∈ B\overline{ρ}(f(x), r) ⊂ V Thus (b_{1}, b_{2}, ...) ∈ R^{ω} is in the same path component as (0, 0, ....) Hence {(b_{1}, b_{2}, ...) ∈ R^{ω} | there exists an M such that |b_{i}| ≤ M} is path connected and hence connected. By our earlier talk in my office this is the largest connected set containing (0, 0, ...) and hence is a component of R^{ω}. Define g_{y} : R^{ω} → R^{ω} by g_{y}(x) = x + y. g_{y}^{−1} : R^{ω} → R^{ω} by g_{y}^{−1}(x) = x − y. Hence g is bijective. g(B\overline{ρ}(x, r)) = B\overline{ρ}(x + y, r) and g^{−1}(B\overline{ρ}(x, r)) = B\overline{ρ}(x − y, r) Thus g and g^{−1} are continuous. C = {(b_{1}, b_{2}, ...) ∈ R^{ω} | there exists an M such that |b_{i}| ≤ M} is connected. Hence g(C) = {(b_{1} + y_{1}, b_{2} + y_{2}, ...) ∈ R^{ω} | there exists an M such that |b_{i}| ≤ M} is connected. Suppose g(C) ⊂ A where A is connected. Then g^{−1} cont implies g^{−1}(A) is connected. Since C ⊂ g^{−1}(A) and C is the largest connected set containing C, C = g^{−1}(A). Thus g(C) = A since g is a bijection. Thus g(C) is also a component. Note x ∈ g(C) iff {x_{1} − y_{1}, x_{2} − y_{2}, ....} is bounded. Thus x and y lie in the same component iff {x_{1} − y_{1}, x_{2} − y_{2}, ....} is bounded.