finding limit of a sequence of vectors

Question

Can someone help me with this limit problem:

Using the definition of a limit in one and multiple dimensions, prove that if $\vec{a}_n \in \mathbb{R}^d$ is a sequence of vectors and $\vec{a} \in \mathbb{R}^d$ is a vector, then $\vec{a}_n \to \vec{a}$ if and only if $\left(\vec{a}_n\right)_i \to a_i$ for all $i=1, \ldots, d$, where $\left(\vec{a}_n\right)_i$ is the $i^{\mathrm{th}}$ coordinate of $\vec{a}_n$ and $\vec{a}_i$ is the $i^{\mathrm{th}}$ coordinate of $\vec{a}$.

Answer 1

HINTS

You need to prove 2 things.

1. If limit exists in each dimension, it also exists in $\mathbb{R}^n$. This can be shown directly -- if in each dimension, the error is small enough, you can bound the norm of the entire error vector.
2. If the limit exists in $\mathbb{R}^n$, it also must exist in each dimension. Here, I would go by contradiction, so assume limit in $\mathbb{R}^n$ exists but in some (WLOG the first) coordinate it does not converge to the needed value. Then, show the error cannot be bounded as the $n$-dimensional definition requires, achieving a contradiction.

Answer 2

In general, suppose that $(X,d_X)$ and $(Y,d_Y)$ are two metric spaces. The question is if we can define a metric $d_{X\times Y}$ on $X \times Y$ such that $\lim_{n\to\infty}(x_n,y_n)=(\lim_{n\to\infty}x_n,\lim_{n\to\infty}y_n)$. So, the idea of passing the limit inside the ordered pair is the essence of defining a metric on the Cartesian product space.

It turns out that we can achieve this in several ways. The most famous product metrics are the ones defined using the $\mathcal{L}^p$ norm in the following way:

$$d_{X\times Y}((x_1,y_1),(x_2,y_2))=\sqrt[p]{d_X(x_1,x_2)^p+d_Y(y_1,y_2)^p}$$

In particular, your case is when $p=2$. Namely, the Euclidean metric. I'm going to show you why this definition works the way we wanted it to work only for the case when $p=2$, but you can do the same for other $1 \leq p < \infty$ nearly the same way. It doesn't hurt to know that when $p=\infty$, the $\mathcal{L}^p$ norm becomes the maximum norm. But it's not necessary to know this now.

Assuming that $x_n \to x$ and $y_n \to y$, it's easy to see that we have:

$$d((x_n,y_n),(x,y)) = \sqrt{d_X(x_1,x)^2+d_Y(y_1,y)^2} \leq d_X(x_n,x)+d_Y(y_n,y) \to 0$$

Hence, $(x_n,y_n) \to (x,y)$.

Conversely, if $(x_n,y_n) \to (x,y)$, it's easy to see that: $$d_X(x_n,x) \leq \sqrt{d_X(x_1,x)^2+d_Y(y_1,y)^2}=d((x_n,y_n),(x,y)) \to 0$$ $$d_Y(y_n,y) \leq \sqrt{d_X(x_1,x)^2+d_Y(y_1,y)^2}=d((x_n,y_n),(x,y)) \to 0$$

Hence, both $x_n \to x$ and $y_n \to y$ are true.