# 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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I've tried to edit your question, but I've left the original version, since at several places I wasn't sure. (I've edited it to what was my best guess.) For example is R1 and Rinfinity in you original version the same thing? \\ To me it seems that you simply copied text from a pdf-file without any editing. If this file is available online, linking to it would add more context for people who would potentially answer your question. –  Martin Sleziak Jan 25 '12 at 9:47
The actual source of the problem is an assignment due Jan 27 in MATH 141 at U C Berkeley: math.berkeley.edu/~jdahl/141/141hw1_2012probs.pdf –  Byron Schmuland Jan 26 '12 at 13:31
@Martin I'm also not sure how to handle this, but the same user has been posting other homework problems (math.stackexchange.com/questions/101470/…) so I'm a bit uneasy. –  Byron Schmuland Jan 26 '12 at 14:30
Dear Buddy Holly: it seems that you are developing a history of asking for help on "textbook exercises" (with timing which unfortunately coincides with homework exercises at Berkeley). Regardless of whether these were actually assigned to you as homeworks, you may find our FAQ on homework questions useful. The community generally dislikes it when you take a tone of voice as if you are assigning us homework. Cheers. –  Willie Wong Jan 26 '12 at 14:54
I've deleted my answer, I will undelete it after Friday. I guess there's no harm in having the solution here after that date; even if it was a homework question. My apologies to the MSE community if I did not follow the guidelines for homework question. Although it did sound homework-like, it wasn't tagged homework, so I did not treat it as such. –  Martin Sleziak Jan 26 '12 at 15:45

## 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