# How to prove the set of bounded sequences is clopen in the uniform metric?

Consider the metric $$d(x,y)=\sup\{\overline d(x_i,y_i)|i\in I\}$$ on $$X=\mathbb R^\omega$$ where $$\overline d(x,y)=\min\{d(x,y),1\}$$ is the standard bounded metric. Consider the topology induced by that metric (the topology whose basis are all open balls in this metric). I'm trying to understand why the subset of all bounded sequences is clopen with respect to this topology.

To show it's open, I need to show that any point (any bounded sequence) has a neighborhood (a union of open balls containing the point) consisting entirely of bounded sequences. I don't have any intuition on how to work with this neighborhood or how to imagine it. Formally, as I said, it's a union of balls, and the ball of radius $$r$$ centered at $$x$$ is $$\{y\in X:d(x,y)< r\}=\{y\in X:\sup\{\overline d(x_i,y_i)|i\in I\} < r\}=\{y\in X: \overline d(x_i,y_i) < r \text{ for all } i\}$$ where the last inequality holds since if the supremum is less than something, then all elements are less than that something. Continuing the above sequence of equalities, $$=\{y\in X:\min\{d(x_i,y_i),1\}< r \text{ for all } i\}$$

How should I think of this ball? How can I see that some appropriate union of such balls that contains a bounded sequence cannot contain unbounded sequences? How to find that union (should it be just a ball centered at the sequence)? The part with closeness is also not clear.

• Work with the complement of your metric space which should be easier – Piquito Sep 26 '18 at 17:01
• I'm not sure why it's easier. In fact, I thought that proving that the subspace is closed is easier by proving that it's complement is open, but for now I don't even see why the subset is itself open, not to mention its complement. – user531587 Sep 26 '18 at 17:22
• The standard bounded metric is $\overline{d}(x,y) = \min(|x-y|,1)$ for $x,y \in \mathbb{R}$. We take the supremum of those values over all coordinates. – Henno Brandsma Sep 26 '18 at 20:23
• You show openness by showing all members to be interior points. Closed is open complement. – Henno Brandsma Sep 26 '18 at 20:30

Let $$B$$ be the set of bounded sequences. Suppose that $$x = (x_n)_n$$ is in $$B$$.
This means that there is some bound $$B$$ such that for all $$n$$: $$|x_n|\le B$$.
If $$y$$ is another sequence in $$\mathbb{R}^\omega$$ such that $$d(x,y) < 1$$, then all $$|y_n -x_n| < 1$$ and this implies that $$y=(y_n)$$ is bounded by $$B+1$$. So the $$d$$-open ball around. $$x$$ with radius $$1$$ sits inside $$B$$ and as $$x \in B$$ was arbitary, $$B$$ is open.
Closedness is similar: if we look at $$U$$, the unbounded sequences, the complement of $$B$$, again a sequence $$y$$ with $$d(x,y) < 1$$ obeys $$|x_n - y_n| < 1$$ for all $$n$$ and so is also unbounded, as $$x$$ is. So also all points of $$U$$ are interior points, $$U$$ is open and so $$B$$ is closed.