Section 20: Problem 5 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let $\mathbb{R}^{\infty}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all sequences that are eventually zero. What is the closure of $\mathbb{R}^{\infty}$ in $\mathbb{R}^{\omega}$ in the uniform topology? Justify your answer.

The finer is the topology on a set, the smaller (at least, not larger) is the closure of any its subset. Indeed, a finer topology has more closed sets, so the intersection of all closed sets containing a given subset is, in general, smaller in a finer topology than in a coarser topology. Another way to see this is that if a point is not in the closure in a coarser topology, then it has a neighborhood that does not intersect the subset, and the same neighborhood will work in a finer topology, so the point is still not in the closure. Yet another way to remember this is that in the finest topology, the discrete topology, the closure of any subset equals the subset itself, while in the coarsest topology, the indiscrete topology, the closure of any nonempty subset is the whole space.

So, in our case, we would expect the closure of $\mathbb{R}^{\infty}$ to be larger (or at least not smaller) than in the box topology, and smaller (or at least not larger) than in the product topology. Exercise 7 of §19 shows that the closure of $\mathbb{R}^{\infty}$ in the box topology is the set itself, and in the product topology is the whole space. So here the answer can be either one of those or anything in between (which does not, actually, help :) ).

Let $X\in\mathbb{R}^{\omega}$ be the set of all sequences of real numbers that converge to $0$ in $\mathbb{R}$ ( $\mathbf{x}\in X$ iff for every $\epsilon>0$ there is $N\in\mathbb{Z}_{+}$ such that for $n\ge N$ , $|x_{n}|<\epsilon$ ). Note, that $\mathbb{R}^{\infty}\subset X$ . If $\mathbf{y}\notin X$ , then there is $\epsilon>0$ such that for every $k\in\mathbb{Z}_{+}$ there is $n_{k}\ge k$ such that $|y_{n_{k}}|\ge\epsilon$ . Hence, if $\mathbf{z}\in B_{\overline{\rho}}(\mathbf{y},\tfrac{\epsilon}{2})$ , for every $k\in\mathbb{Z}_{+}$ , $|z_{n_{k}}|>|y_{n_{k}}|-\tfrac{\epsilon}{2}\ge\tfrac{\epsilon}{2}$ , and $B_{\overline{\rho}}(\mathbf{y},\tfrac{\epsilon}{2})$ does not contain any points of $X$ . Therefore, $X$ is closed and contains the closure of $\mathbb{R}^{\infty}$ . At the same time, for every $\mathbf{x}\in X$ and $\epsilon>0$ , there is $N\in\mathbb{Z}_{+}$ such that for $n\ge N$ , $|x_{n}|<\tfrac{\epsilon}{2}$ , and $\mathbf{y}=(x_{1},\ldots,x_{N},0,0,\ldots)\in$ $B_{\overline{\rho}}(\mathbf{x},\epsilon)\cap\mathbb{R}^{\infty}$ . So, the closure of $\mathbb{R}^{\infty}$ in the uniform topology is the set of all sequences of real numbers converging to zero in $\mathbb{R}$ .

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Section 20: Problem 5 Solution

Section 20: Problem 5 Solution