1

I have a question regarding Theorem 20.4 in Munkres' topology. The theorem states that the uniform topology on $\mathbb{R}^J$ is finer than the product topology and coarser than the box topology.

The uniform topology is defined as $\bar{\rho}(x,y) = \mathrm{sup}\{\bar{d}(x_{\alpha}, y_{\alpha})|\alpha \in J\}$ where $\bar{d}$ is the standard bounded metric.

My confusion arises in showing that the box topology is finer than the uniform topology. The proof states that we should let $B$ be the $\epsilon$-ball centered at $x$ in the $\bar{\rho}$-metric. Then the box neighborhood $$U = \prod (x_\alpha - \frac{1}{2}\epsilon, x_{\alpha} + \frac{1}{2}\epsilon)$$ of $x$ is contained in $B$, for if $y \in U$, then $\bar{d}(x_\alpha, y_\alpha) < \frac{1}{2}\epsilon, \space \forall \alpha$, so that $\bar{\rho}(x,y) \leq \epsilon/2$.

It is the last inequality that I don't understand. Or rather, the fact that it is not a strict inequality. If $\bar{d}(x_\alpha, y_\alpha)$ is strictly smaller than $\epsilon/2$ for all $\alpha$, then how could $\bar{\rho}(x,y)$ ever be equal to $\epsilon/2$?

frej.mh
  • 193

2 Answers2

2

If $x = (0,0,0,\ldots)$ is one point (index over $\mathbb{N}$) and $y = (0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots, \frac{n-1}{n}, \ldots)$, then $\bar{d}(x_n, y_n) < 1$ for all $n$, but the supremum of their differences is $1$,a s $\frac{n-1}{n} \to 1$, so we have $\bar{\rho}(x,y) = 1$, not $<1$.

If all $\bar{d}(x_\alpha, y_\alpha) < \frac{\varepsilon}{2}$ we know that $\frac{\varepsilon}{2}$ is an upper bound for the set $\{\bar{d}(x_\alpha, y_\alpha), \alpha \in J\}$ and $\bar{\rho}(x,y)$ is the least upper bound of that set, hence $\bar{\rho}(x,y) \le {\varepsilon \over 2}$ and the above example shows we cannot do better, a set can have a supremum larger than all the elements in the set, as $\sup(0,1) =1$ already shows.

Henno Brandsma
  • 242,131
  • +1 very nice explanation @ Henno sir pliz answer my post https://math.stackexchange.com/questions/2770067/1-why-u-mathbfx-epsilon-is-open-in-box-product-2-why-u-mathbfx –  May 07 '18 at 03:53
0

For finite J they are the same. Is the open box set PI_n U_n where
U_2k = (0,1), U_(2k-1) = R, k in N, open in the other two topologies?

William Elliot
  • 17,598