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I would like to show the Bolzano-Weierstrass in $\mathbb{R}^n$, I have seen this theorem in $\mathbb{R}$ and I know it can be shown by induction, something I will try now.

Theorem : Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence in $\mathbb{R}^n$

Proof : By induction : we know that it is true for $n=1$, we assume this holds for $p=k$ and we want to show this holds for $p=k+1$. Consider $(x_n)$ a bounded sequence in $\mathbb{R}^{k+1}$. For all $n\in\mathbb{N}$ we can write $x_n = (a_n, x_{n}^{k+1})$ where $a_n = (x_{n}^{1}, ..., x_{n}^{k})\in\mathbb{R}^{k}$ since $\mathbb{R}^{k+1}$ is isomorphic to $\mathbb{R}^{k}\times\mathbb{R}$ so this rewritting of $x_n$, even if not exactly the same "element" as the initial $x_n$, can be treated as the same.

Clearly, for all $n\in\mathbb{N}, \exists M>0 : \lvert x_{n}^{k+1}\rvert\leq\lVert x_n\rVert_{2}\leq M$ and $\lVert a_n\rVert_{2}\leq\lVert x_n\rVert_{2}\leq M$.

Using the induction hypothesis, we know that $(a_n)$ has a convergent subsquence $(a_{n_j})$, we denote $x\in\mathbb{R}^{k}$ its limit. Now, the sequence $(x_{n}^{k+1})$ has also a convergent subsequence in $\mathbb{R}$ by Bolzano-Weierstrass with limit $x^{k+1}\in\mathbb{R}$. This shows that $x_n$ admits a convergent subsequence $(a_{n_j}, x_{n_j}^{k+1})$ whose limit is $(x,x^{k+1}) = (x^1, ..., x^{k+1})\in\mathbb{R}^{k+1}$ so the Bolzano-Weierstrass is true for $p=k+1$, which concludes the proof.

This seems correct to you?

EDIT : My proof is false, we need to take another subsequence to be sure to have a vector with coordinate that have the same index ! Thanks to FShrike, Vercassivelaunos and egreg, you will find three very clear answers to this problem below.

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    $\begingroup$ No, it is not correct. From the fact $(a_n)_{n\in\Bbb N}$ has a convergent subsequence and $(x_n^{k+1})_{n\in\Bbb N}$ also has a convergent subsequence you cannot deduce that $(x_n)_{n\in\Bbb N}$ converges (which, by the way, is stronger than what you were aiming at). $\endgroup$ Commented Dec 26, 2022 at 14:28
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    $\begingroup$ Just because the subsequence $\left( a_{n_j}\right)$ converges does not mean the sequence $(a_n)$ will converge to $x$. The same is true for $(x_n^{k+1})$. That sequence doesn't necessarily converge either. $\endgroup$
    – Michael
    Commented Dec 26, 2022 at 14:30
  • $\begingroup$ Thank you José and Michael for your comments. Indeed my conclusion was not the one I wanted, I edited this. This should be better now $\endgroup$
    – G2MWF
    Commented Dec 26, 2022 at 14:44
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    $\begingroup$ Unfortunately the changes are also incorrect. You know $(x_n^{k+1})$ will have a convergent subsequence, but you don't know that the indices of that subsequence will be the same as those for $(a_{n_j})$. That is, $(x_n^{k+1})$ will have a convergent subsequence, say $(x_{r_j}^{k+1})$. Thus the sequence $(a_{n_j}, x_{r_j}^{k+1})$ will certainly converge, but this is not necessarily a subsequence of $(x_n)$. $\endgroup$
    – Michael
    Commented Dec 26, 2022 at 14:48
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    $\begingroup$ More can be proved: let $\{x_n^{(k)}\}_{n=1}^\infty $ $k=1,2,\ldots,$ be an infinite family of bounded sequences. There exists an increasing subsequence $n_j$ of positive integers so that the sequence $\{x_{n_j}^{(k)}\}_{j=1}^\infty$ is convergent for every $k.$ In your case we are dealing with finitely many sequences. $\endgroup$ Commented Dec 26, 2022 at 16:58

3 Answers 3

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You can fix this without too much bother. Heed the comments well, because they gave you very important warnings. You must make sure you are using the same subsequence for both components, if you want to take a genuine subsequence of $(x_n)_n$ in $\Bbb R^{k+1}$.

By the ordinary BW theorem, there is an increasing sequence $(n_j)_j$ with $(x^{(k+1)}_{n_j})_j$ convergent. Consider the subsequence $(a_{n_j})_j$ which is still a bounded sequence in $\Bbb R^k$. Using the inductive hypothesis, there is a further subsequence $(n_{j_i})_i$ along which $(a_{n_{j_i}})_i$ is convergent. If this is confusing, declare the sequence $b_j:=a_{n_j}$ in $\Bbb R^k$ and note $(b_j)_j$ is a bounded sequence: by induction, it has a convergent subsequence $(b_{j_i})_i$, which translates to $(a_{n_{j_i}})_i$.

Then, because subsequences of a convergent sequence are convergent, we know that $(x^{(k+1)}_{n_{j_i}})_i$ is convergent. It now follows that $(x_{n_{j_i}})_i$ is a convergent subsequence in $\Bbb R^{k+1}$ since this subsequence is convergent in both the $a$ and $x^{(k+1)}$ components.


If you know about this sort of thing, you might like to learn that Bolzano-Weierstrass is true in any complete metric space where bounded subsets are always totally bounded. Sequential compactness (essentially this is Bolzano-Weierstrass) is equivalent to compactness which is further (generalised Heine-Borel) equivalent to completeness and total boundedness (in Euclidean space, that is just closed and bounded).

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    $\begingroup$ Thank you a lot for your answer ! $\endgroup$
    – G2MWF
    Commented Dec 26, 2022 at 15:22
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    $\begingroup$ @coboy you are most welcome $\endgroup$
    – FShrike
    Commented Dec 26, 2022 at 15:23
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You have to go one layer deeper with your subsequences. You know the following:

  • Any bounded sequence in $\mathbb R$ has a convergent subsequence.
  • Any bounded sequence in $\mathbb R^k$ has a convergent subsequence.
  • The sequence $(a_n,x_n^1,\dots,x_n^k)$ is a bounded [convergent] sequence in $\mathbb R^{k+1}$ if and only if $(a_n)$ and $(x_n^1,\dots,x_n^k)$ are bounded [convergent] sequences in $\mathbb R$ and $\mathbb R^k$, respectively.

Now to show that $(a_n,x_n^1,\dots,x_n^k)$ has a convergent subsequence if it is bounded, we first have to consider the $a_n$ component. This is also bounded, and thus has a convergent subsequence $a_{n_i}$. Now consider the subsequence $(a_{n_i},x_{n_i}^1,\dots,x_{n_i}^k)$. Its $x_{n_i}$ component is a subsequence of $(x_n^1,\dots,x_n^k)$, not necessarily convergent, but bounded. So it has a convergent subsequence $(x_{n_{i_j}}^1,\dots,x_{n_{i_j}}^k)$. Now, the corresponding subsequence $a_{n_{i_j}}$ is a subsequence of the convergent subsequence $a_{n_i}$, and is thus itself convergent.

So taken together, the two convergent "subsubsequences" yield a convergent subsequence $(a_{n_{i_j}},x_{n_{i_j}}^1,\dots,x_{n_{i_j}}^k)$ of our original sequence.

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    $\begingroup$ Dear Vercassivelaunos, thank you a lot. I just understood thanks to your answer what was wrong with my proof, thank you very much. $\endgroup$
    – G2MWF
    Commented Dec 26, 2022 at 15:22
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You don't really need induction, although it makes things simpler. A sequence in $\mathbb{R}^{k+1}$ can be written in the form $(a_n,b_n)$, where $a_n\in\mathbb{R}^{k}$ and $b_n\in\mathbb{R}$.

Suppose $(a_n,b_n)$ is bounded, so both $(a_n)$ and $(b_n)$ are bounded.

By the induction hypothesis we find a convergent subsequence from $(a_n)$, let it be $(a_{n_m})$. Consider $(b_{n_m})$, which is a bounded sequence, so it has a convergent subsequence $(b_{n_{m_l}})$. OK, the notation becomes cumbersome, but we're done, because also $(a_{n_{m_l}})$ is convergent and therefore also the combined subsequence $(a_{n_{m_l}},b_{n_{m_l}})$ in $\mathbb{R}^{k+1}$ is convergent.

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