Section 19: Problem 7 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," that is, all sequences $(x_{1},x_{2},\ldots)$ such that $x_{i}\neq0$ for only finitely many values of $i$ . What is the closure of $\mathbb{R}^{\infty}$ in $\mathbb{R}^{\omega}$ in the box and product topologies? 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) in the product topology than in the box topology.

If $\mathbb{R}^{\omega}$ is given the product topology, for every point $\mathbf{x}\in\mathbb{R}^{\omega}$ and every its neighborhood $U=\prod_{n\in\mathbb{Z}_{+}}U_{n}$ , where for all but finitely many values of $n$ , $U_{n}=\mathbb{R}$ , choose a point $\mathbf{y}$ such that $y_{n}\in U_{n}$ and $y_{n}=0$ if $U_{n}=\mathbb{R}$ . Since for only finitely many values of $n$ , $U_{n}$ is a proper subset of $\mathbb{R}$ , $\mathbf{y}\in\mathbb{R}^{\infty}$ , hence, $\mathbf{y}\in U\cap\mathbb{R}^{\infty}$ . We conclude that $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}$ in the product topology (see also Example 7 of §23).

An alternative way to see that $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}$ is as follows. According to Exercise 6, a sequence in the product space converges to a point iff all coordinate projections converge to the corresponding projections of the point. For $\mathbf{x}=(x_{1},x_{2},...)$ in $\mathbb{R}^{\omega}$ define a sequence of points $\mathbf{y}_{k}\in\mathbb{R}^{\infty}$ such that the first $k$ coordinates of $\mathbf{y}_{k}$ are equal to the first $k$ coordinates of $\mathbf{x}$ , and all others are zero. Then, clearly, the projections of $\mathbf{y}_{k}$ ’s onto any coordinate $n$ converge to $x_{n}$ , and, therefore, the sequence $\mathbf{y}_{1},\mathbf{y}_{2},\ldots$ converges to $\mathbf{x}$ , i.e. every $\mathbf{x}$ is in the closure of $\mathbb{R}^{\infty}$ in the product topology.

Now, if $\mathbb{R}^{\omega}$ is given the box topology, every point not in $\mathbb{R}^{\infty}$ has a disjoint open neighborhood. Indeed, in the box topology, if a point $(x_{1},x_{2},...)\notin\mathbb{R}^{\infty}$ then there is an increasing infinite sequence $\{n_{k}\}$ of indexes such that $x_{n_{k}}\neq0$ for all $k$ , and for each $k$ there is a neighborhood $U_{n_{k}}$ of $x_{n_{k}}$ that does not contain $0$ . For all other coordinates $m$ (not in $\{n_{k}\}$ ) let us take $U_{m}=\mathbb{R}$ . Then, $\prod_{n}U_{n}$ is open in the box topology, contains $\mathbf{x}$ , but does not intersect $\mathbb{R}^{\infty}$ . Therefore, $\mathbb{R}^{\omega}-\mathbb{R}^{\infty}$ is open, and $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$ in the box topology.

Subpages

Section 19: Problem 7 Solution

Section 19: Problem 7 Solution