# Proving a set is complete

I am working on this problem for an introductory Real Analysis course.

Let $\mathbf{x_0} \in \mathbb{R}^n$ and $R > 0$. Prove that $B = \{ \mathbf{x} \in \mathbb{R}^n : || \mathbf{x} - \mathbf{x_0}|| \leq R \}$ is complete.

Let $(\mathbf{b}_n)^{\infty}_{n=1}$ be any Cauchy sequence in $B$, then $(\mathbf{b}_1,\mathbf{b}_2, \mathbf{b}_3,....,\mathbf{b}_n,... ) \in B$ and as such

$$||\mathbf{b}_i - \mathbf{x_0}|| \leq R$$

for all $i \in \mathbb{I}$ and $i>0$. Further, $\mathbf{b}_n$ is Cauchy and as such must converge to a point $\mathbf{a}$, with

$$\lim_{n \to \infty} \mathbf{b}_n = \mathbf{a}.$$

Now, I believe I need to show that $\mathbf{a} \in B$. I am unsure about this part. I take

$$\lim_{n \to \infty} ||\mathbf{b}_n - \mathbf{x_0}|| \leq \lim_{n \to \infty} R$$ $$|| \lim_{n \to \infty} \mathbf{b}_n - \mathbf{x_0}|| \leq R$$ $$|| \mathbf{a} - \mathbf{x_0}|| \leq R.$$

Showing that the limit point is contained in $B$. But then couldn't I use the same argument for a set $A$ with the same properties as $B$ but $|| \mathbf{x} - \mathbf{x_0}|| < R$? Then the open set would contain all of its limit points. I guess what I am struggling with is showing that any sort of closed interval contains its limit points.

I realize that $B$ is a closed subset of $\mathbb{R}^n$, and as such contains all of its limit points. However, in my textbook (Real Analysis and Applications, Davidson and Donsig), the concept of a closed set is not covered until the next section. As such, I am hoping to complete this proof without the notion of a closed set, or compactness, or balls, et cetera.

I was hoping for a proof that would show that you can take any Cauchy sequence in $B$, show that it converges to a point in $B$, and as such show that $B$ is complete. But it seems to me simply by stating that the sequence is in $B$ means it should converge to a point in $B$ (when would it not?). Any help or direction would be appreciated.

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Consider $B = (0,1) \subseteq \mathbb{R}$. The sequence $x_n = 1/n$ is Cauchy, every term lives in $B$, but the limit $\lim_{n\to \infty} x_n = 0\not\in B$. – Willie Wong Feb 26 '13 at 16:12
A simple generalisation of the previous comment shows that $\|b_n - x_0\| < R ~~~\not\!\!\!\!\implies \lim_{n\to\infty}\|b_n - x_0\| < R$. So no, you cannot do it "for the set $A$". – Willie Wong Feb 26 '13 at 16:15

Let us first prove that $\mathbb{R}^n$ is complete(you may skip reading this part if your text book has already done this). Let $\{x_1 , \ldots x_n, \ldots \}$ be a cauchy sequence in $\mathbb{R}^n$ (I trust you understand my notation, as I prefer not to write elements of $\mathbb{R}^n$ in bold). Let $x_i^{(j)}$ be the j'th component of the vector $x_i$. Convince yourself that $|x_n^{(j)} - x_m^{(j)}| \le \|x_n - x_m \|$ for all $m,n,j$. Now we fix $j$. Since $\{x_i\}_i$ is a Cauchy sequence we claim that $\{x_n^{(j)} \}$ is also Cauchy. This is not very hard to show at given $\epsilon > 0$ we can find $N$ such that whenever $n,m \ge N$ we have $\|x_n - x_m \| < \epsilon$ but by the inequality above we must also have $|x_n^{j} - x_m^{j}| < \epsilon$ Hence $x_n^{(j)}$ is also Cauchy as claimed. Now since we know that $\mathbb{R}$ is complete it follows that $x_n^{j}$ converges to some real number which we denote $x^{(j)}$. Since $j$ is arbitrary this holds for all $j$. Now We claim that the Cauchy sequence $\{x_i \}$ converges to $x = (x^{(1)},x^{(2)}, \ldots , x^{(n)})$. Given $\epsilon > 0$ we have $\|x_n - x \| = \|(x_n^{(1)}-x^{(1)})e_1 + \ldots (x_n^{(n)} - x^{(n)})e_n \| \le |x_n^{(1)} - x^{(1)}| + \ldots |x_n^{(n)} - x^{(n)}|$. Where $e_i = (0, \ldots 1 , \ldots 0)$ where the $1$ is at the i'th component. Now by construction $x_n^{(j)}$ converges to $x^{(j)}$ Hence we can find $N_i$ such that whenever $n \ge N_i$ we have $|x_n^{(i)} - x^{(i)}| < \epsilon/n$. Letting $N = max(N_i)$ we see that $|x_n^{(1)} - x^{(1)}| + \ldots |x_n^{(n)} - x^{(n)}| < \epsilon$ whenever $n \ge N$. Hence the Cauchy sequence $x_n$ converges to $x$ as claimed and we have proved that $\mathbb{R}^n$ is complete.

Now to answer your question: Like you said we can assume that $b_n \to a$. We must show that $a \in b$. Assume that $\|a - x_0 \| > R$ then $\| a - x_0 \| = R + k$ where $k > 0$. We have $\|a-x_0\| \le \|a - b_n \| + \|b_n-x_0\| \le \|a-b_n\| + R$ but this is a contradiction since we can find an $N$ such that $\|a - b_N \| < \epsilon = k$ since $b_n \to a$. Q.E.D.

Final remark: You see that this argument doesn't work if we had $\| a - x_0 \| = R$ as then $k = 0$ and you can actually see a good counter example given by the comment above by Willie Wong.

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From the triangle inequality you have $\|\mathbf{a}-\mathbf{x}_0\|\le\|\mathbf{a}-\mathbf{b}_n\|+\|\mathbf{b}_n-\mathbf{x}_0\|\le\|\mathbf{a}-\mathbf{b}_n\|+R$ for all $n\in\Bbb N$. You know that $\lim_{n\to\infty}\|\mathbf{a}-\mathbf{b}_n\|=0$, so $\|\mathbf{a}-\mathbf{x}_0\|\le R$, and $\mathbf{a}\in B$.

This wouldn’t work if $B$ were the open ball of radius $R$, because from $\|\mathbf{a}-\mathbf{x}_0\|\le\|\mathbf{a}-\mathbf{b}_n\|+R$ for all $n\in\Bbb N$ you can still conclude only that $\|\mathbf{a}-\mathbf{x}_0\|\le R$, not that $\|\mathbf{a}-\mathbf{x}_0\|<R$.

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