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I apologize for this very basic definition question, but I am stumped because my professor uses this notation without any introduction.

Verbatim,

"Recall that one of the properties of algebraic convergence is that $(X_n, Y_n) \rightarrow (X,Y)$ iff $X_n \rightarrow X$ and $Y_n \rightarrow Y$."

What does $(X_n, Y_n)$ mean? Google results show that $(X_n, Y_n)$ may mean an ordered pair. But then what is the definition of the convergence of an ordered pair?

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For a given $n \in \mathbb{N}$, $(X_n,Y_n)$ is an ordered pair. When $n$ varies, though, you get a sequence which can be denoted by $\{(X_n,Y_n)\}_{n=0}^{\infty}$.

"It is said to converge if and only if each component sequences $\{X_n\}_{n=0}^{\infty}$ and $\{Y_n\}_{n=0}^{\infty}$ converges in the usual sense."

This is actually a theorem if you define convergence in $\mathbb{R^2}$ using the distance $d:\mathbb{R}^2\to\mathbb{R}$ given by the square root of the difference of the squares of each component.

Given a sequence $\{A_n\}_{n=0}^{\infty}=\{(X_n,Y_n)\}_{n=0}^{\infty}$ with $X_n,Y_n \in \mathbb{R}$ for each $n\in \mathbb{N}$ we say it is a convergent sequence if there is an $C=(C_1,C_2) \in \mathbb{R^2}$ (a limit) and for every $\varepsilon >0$ (tolerance) there is a $N\in \mathbb{N}$ such that for every $n>N$ we have $d(C,A)<\varepsilon$

Roughly speaking, if you know a sequence converges you get a $N$, you can then use it to show that each of the component sequence has an $N_x$ and a $N_y$ that work for each of their own convergence. The other direction is similar.

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  • $\begingroup$ So the sentence that I quote IS the definition of ordered pair's convergence? But then I'm supposed to prove that $(X_n, Y_n) \rightarrow^p (X,Y)$ iff $X_n \rightarrow^p X$ and $Y_n \rightarrow^p Y$. How do I now define $(X_n, Y_n) \rightarrow^p (X,Y)$ then? $\endgroup$ – Heisenberg Sep 29 '13 at 3:27
  • $\begingroup$ n this case the definition is just like the usual definition for sequences, except you use distances instead of absolute values. I'm going to clarify it. $\endgroup$ – user89138 Sep 29 '13 at 3:32
  • $\begingroup$ I asked my professor today, and he claimed that $(X_n, Y_n)$ is actually a vector, with $X_n$ and $Y_n$ also being vectors themselves (possibly of different dimensions). Could you please explain? $\endgroup$ – Heisenberg Oct 1 '13 at 20:22
  • $\begingroup$ Well it all depends on the context. You can define them to be whatever you want them to be. In his class I'm guessing you can atribute numbers to $X_n$ and $Y_n$, so the pair $(X_n,Y_n)$ can be viewed as a vector on a plane. But, if you want to, you can define each of those as vectors themselves so that you actually get a pair of vetors instead of numbers. $\endgroup$ – user89138 Oct 2 '13 at 5:05

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