Coordinate-Wise Convergence in R^n .... TB&B Chapter 11, Section 11.4 ....

[Math Amateur]

I am reading the book "Elementary Real Analysis" (Second Edition, 2008) Volume II by Brian S. Thomson, Judith B. Bruckner, and Andrew M. Bruckner ... and am currently focused on Chapter 11, The Euclidean Spaces \(\displaystyle \mathbb{R}^n\) ... ...

I need with the proof of Theorem 11.15 on coordinate-wise convergence of sequences in \(\displaystyle \mathbb{R}^n\) ... note that the proof precedes the theorem statement in TB&B ... ...

Theorem 11.15 and its proof (as preceding notes) reads as follows:View attachment 7711
View attachment 7712Can someone please explain exactly how (8) follows from (7) ...Help will be much appreciated ...

Peter

 

[GJA]

Hi, Peter.

Define the vector $y=x_{k}-x$, then apply $(7)$ to $y$ to get $(8)$.

 

[math771]

Hi Peter,

I would be happy to help explain the proof of Theorem 11.15 for you. The theorem states that if a sequence of points in \mathbb{R}^n converges coordinate-wise, then it also converges in the Euclidean metric. The proof, as stated in TB&B, goes as follows:

First, we assume that the sequence (x_k) converges coordinate-wise to some point x = (x_1, ..., x_n). This means that for each i = 1, ..., n, the sequence (x_k^i) converges to x_i.

Next, we consider the Euclidean metric d(x_k, x). This is defined as d(x_k, x) = \sqrt{\sum_{i=1}^{n} (x_k^i - x_i)^2}.

Now, since each coordinate sequence (x_k^i) converges to x_i, we can choose a large enough k such that for all i = 1, ..., n, we have |x_k^i - x_i| < \epsilon, where \epsilon > 0 is some small number.

Using this, we can show that d(x_k, x) < \sqrt{n\epsilon^2} = \epsilon\sqrt{n}. This follows from the fact that d(x_k, x) is a sum of squares, and each square is less than or equal to \epsilon^2. Therefore, the total sum is less than or equal to n\epsilon^2.

Finally, since \epsilon\sqrt{n} is arbitrary, we can choose \epsilon to be as small as we want, which means that d(x_k, x) can be made arbitrarily small. This shows that the sequence (x_k) converges to x in the Euclidean metric, and thus proves Theorem 11.15.

I hope this helps clarify the proof for you. Let me know if you have any further questions.