Usbrth

Show that a sequence $(x_1^n,x_2^n)$ in $mathbbR^ptimesmathbbR^q$ converges to $(x_1,x_2)$ iff the same thing happens when we consider the sequence as belonging to $mathbbR^p+q$.

I am not able to understand the notations used, neither how to proceed.

Here are my beginner questions:-

1. Is $(x_1^n)$ a sequence in $mathbbR^p$ ?


2. Why indexing for sequencing is in superscript, instead of subscript (as is usual) ?


3. Is $x_1^1$ a vector in $mathbbR^p$ ?


4. How can $(x_1^n,x_2^n)$ be a sequence in $mathbbR^p+q$ ? (since it is a pair).

Also, please provide me pointers to begin the proof.

1) Yes.

2) If you want to consider a sequence in $mathbb R^p$ it is natural to use superscripts because each element of the sequence already has coordinates. We use subscripts for the coordinates.

3)Yes

4) If $x=(x_1,x_2,...,x_p) in mathbb R^p$ and $y=(y_1,y_2,...,y_q) inmathbb R^q$ the notation $(x,y)$ is often used as an abbreviation for $(x_1,x_2,...,x_p,y_1,y_2,...,y_q)$ which is an element of $mathbb R^p+q$

For a proof use the following: if $x$ and $y$ are as above then $|(x,y)|=sqrt y$.

Hence the distance between $(x_1^n,y_1^n)$ and $(x,y)$ is $sqrt ^2$ which tends to $0$ iff both $|x_1^n-x|^2$ and $|y_1^n-y|^2$ tend to $0$.

1. Yes, it is.




2. Because they use the subscript to separate the sequence in $Bbb R^p$ from the sequence in $Bbb R^q$. They could have used $(x_n, y_n)$, but they went for $(x_1^n, x_2^n)$. I wouldn't have made that choice myself, but ultimately it's about aesthetics, and has no actual mathematical impact.




3. Yes, it is (this feels like a special case of question 1.)




4. For any $n$, $x_1^n$ is a point in $Bbb R^p$, and as such, it can be characterised by coordinates as a $p$-tuple like $x_1^n = (x_1, 1^n, x_1, 2^n, ldots, x_1, p^n)$ (sorry about the indexing). Similarily, $x_2^n$ is a $q$-tuple. Put one after the other, and you get a $(p+q)$-tuple of coordinates, which is to say a point in $Bbb R^p+q$. Since this is done for each $n$, you get a sequence.