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Then, we can define an inner product over this space as: $\mathbf{x}\cdot \mathbf{y} := \sum x_iy_i$. (This is a finite sum.) Ok, let $X$ the set of vectors $e_i = (0,\ldots,1,0,\ldots,\ldots)$, where the unique entrace that differs from zero is the input $i.$

This is an example of set that is bounded, note that $X \subset B_{2\sqrt{2}}(e_1)$ the open ball centered in $e_1$ and of radius $2\sqrt{2}.$

It is not compact! once for each $i$, $B_{\frac{\sqrt{2}}{2}}(e_i)$ contains only $e_i,$ and that is an open conver that cannot be refined to a finite subcover.

But it is also closed! And tha is what I am not able to prove. Any hint, comment, answer will be very appreciate.

Thanks a lot.

L.F. Cavenaghi
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2 Answers2

1

Let $x = \sum_{i=1}^n \lambda_i e_i \in \mathbb{R}^\infty$.
If $x \in \overline{X}$, we can find a sequence $(e_{i(m)})_{m \in \mathbb{N}}$ such that $\lim_{m \to \infty} \| e_{i(m)} - x \|^2 = 0$. This implies $$0 = \lim_{m \to \infty} \| e_{i(m)} - x \|^2 = \lim_{m \to \infty} \sum_{i \neq i(m)} \lambda_{i(m)}^2 + \left( 1- \lambda_{i(m)} \right)^2.$$

So we need $\lambda_{i} \in \{ 0, 1\}$, else the limit can never be $0$.
Furthermore, we must have $ \sum_{i=1}^n \lambda_i=1$, so exactly one $\lambda_i$ is one and the rest is zero.
Thus $x = e_i \in X$.

And therefore $\overline{X} = X$.

Hetebrij
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To show that it is closed, you can also use the open cover $U_i = B_{2^{-1/2}}(e_i)$ you gave above. Suppose that $(x_n)$ is a sequence in $X$ which converges, say to $y \in \mathbf R^{\infty}$. Then $(x_n)$ is a Cauchy-sequence, that is, there is $N \in \mathbf N$, such that $d(x_n, x_m) < 2^{-1/2}$ for $n,m \ge N$. As $B_{2^{-1/2}}(x_N)$ contains only one element of $X$, we must have $x_m = x_N$ for $m \ge N$. Hence $y = x_N \in X$.

martini
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