# basis closed, bounded

This is Exercise 3 from p. 39 of Munkres: Analysis on Manifolds

Let $\mathbb R^\infty$ be the set of all "infinite-tuples" $x = (x_1, x_2, \ldots )$ of real numbers that end in an infinite string of $0$s. (See the exercises of § 1.)

Define an inner product on $\mathbb R^\infty$ by the rule $\langle x, y\rangle = \sum x_iy_i$. (This is a finite sum, since all but finitely many terms vanish.) Let $\|x - y\|$ be the corresponding metric on $\mathbb R^\infty$. Define $$e_i = (0, \ldots, 0, 1, 0, 0, \ldots);$$ where 1 appears in the i-th place. Then the $e_i$ form a basis for $\mathbb R^\infty$.

Let $X$ be the set of all the points $e_i$. Show that $X$ is closed, bounded, and non-compact.

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## 1 Answer

It should be easy to show, that if $i\ne j$ then $\|e_i-e_j\|=\sqrt2$. Using this fact you can show easily if a sequence of points from $X$ is convergent, then it is eventually constant. Indeed, let $(x_i)$ be a convergent sequence of points of $X$. Then it is Cauchy, i.e. there is $N$ such that $p,q>N$ implies $\|x_p-x_q\|<\sqrt{2}$. This clearly implies that all values $x_p$ for $p>N$ must be the same.

The above fact should help you to show the closedness. (A subset $X$ of a metric space is closed if and only if for every convergent sequence of points of $X$ the limit belongs to $X$ as well, see e.g. proofwiki.)

A different approach: You could try to show that the complement $\mathbb R^\infty\setminus X$ is open. Hint: Start by showing that for $x_0\notin X$ the neighborhood $U(x_0;\frac{\sqrt2}2)$ contains at most one point of $X$.

Boundedness is easy. Just notice that $\|e_i\|=1$. I.e. we have $\|x\|\le 1$ for each $x\in X$. (Definition of boundedness is that $X$ is bounded if and only if there exists an $M$ such that $\|x\|\le M$ for each $x\in X$.)

If a metric space is compact, then every sequence has a convergent subsequence. Again the fact that $\|e_i-e_j\|=\sqrt2$ should suffice you to show that $(e_i)$ has no convergent subsequence. (Recall that every convergent sequence is Cauchy, see e.g. proofwiki.)

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Can you please walk me through the steps, I'm not seeing how boundedness is "easy". All I note is that the norm of e sub i = 1 for all i, so is that all we can say for boundedness? – Buddy Holly Jan 25 '12 at 10:27
What you wrote is sufficient for boundedness. (I've added the part about boundedness to my post.) – Martin Sleziak Jan 25 '12 at 10:31
For closedness, can you elaborate on your argument? I'm not sure that I have the idea down – Buddy Holly Jan 25 '12 at 10:46
I've added more details. Of course, you could also try other approach to show $X$ is closed. (There are many ways to do this.) – Martin Sleziak Jan 25 '12 at 10:56
How do you use the fact fact that you can show easily if a sequence of points from X is convergent, then it is eventually constant? I am just going to say, by the fact that all Cauchy sequences are closed and bounded. Is this sufficient? – Buddy Holly Jan 26 '12 at 10:50