# Bolzano Weierstrass theorem in a finite dimensional normed space

The problem may have a very simple answer, but it is confusing me a bit now.

Let $(\mathbf{V},\lVert\cdot\rVert)$ be a finite dimensional normed vector space. A subset $\mathbf{U}$ of $\mathbf{V}$ is said to be bounded, if there is a real $M$ such that for any member $u$ of $\mathbf{U}$, we have: $\lVert u\rVert\lt M$. . Also, convergence of a sequence in $\mathbf{V}$ is defined with respect to the metric $\lVert\cdot\rVert$. Is it true that every bounded sequence of vectors in $\mathbf{V}$ admits a convergent subsequence?

If not, please give a counterexample with $\mathbf{V}$ finite dimensional.

• Yes, because it holds for every $\mathbb{R}^n$ with any norm (because they are all equivalent). Your space is isomorphic (then homeomorphic) to some $\mathbb{R}^n$, hence the notions of convergence on your $V$ can be viewed as convergence in this $\mathbb{R}^n$ May 13, 2012 at 4:01

I will write a little bit more details about my comment.

Take a basis $\{v_1,\ldots,v_n\}$ for $V$ and consider the isomorphism $T:\mathbb{R}^n\rightarrow V$ such that $T(e_i)=v_i$. Define a new norm on $\mathbb{R}^n$ by

$$\|x\|_{\mathbb{R}^n}=\|T(x)\|_V$$

This implies $T$ is a homeomorphism, because it takes opens balls onto open balls of the same radius, by definition of the norm on $\mathbb{R}^n$. In particular, remember that a homeomorphism takes convergent sequences in convergent sequences.

Once $T$ is a isomorphism, this is in fact a norm and, in addition, we have

$$\|v\|_V=\|T^{-1}(v)\|_{\mathbb{R}^n}$$

Take a bounded sequence $\{v_k\}_{k=1}^\infty$ on $V$. Then $x_k=T^{-1}(v_k)$ is a bounded sequence on $(\mathbb{R}^n,\|\cdot\|_{\mathbb{R}^n})$. Once all the norms on $\mathbb{R}^n$ are equivalent, this sequence is bounded with respect to the euclidean norm. Then, this sequence have a convergent subsequence, say $x_{k_j}$. Hence, $T(x_{k_j})$ is a convergent subsequence of the original $\{v_k\}_{k=1}^\infty$.

• But why does not work in an infinite dimensional real vectors space ? i,e which step of the proof fails in infinite dimensional case ? I mean we can still construct isomorphisms in such cases.
– Our
Nov 4, 2017 at 10:27
• @Our Norms on finite dimensional vector space are equivalent.
– Jiya
Jan 9 at 12:54

Here is a short answer but it depends on other results.

1. Every $$n$$-dimensional normed linear space $$V$$ is complete. It is in fact homeomorphic to $$R^n$$.
2. Recall that a set in $$R^n$$ is compact if and only if it is closed and bounded.
3. Following this, we can show that a set in an $$n$$-dim normed linear space is compact if and only if it is closed and bounded.
4. If a set $$U \subseteq V$$ is bounded. Then its closure is closed and bounded, hence compact.
5. Finally, we use the fact that every sequence of a compact set has a convergent subsequence that converges in the compact set itself.
6. Thus, if $$U$$ is bounded, then every sequence of $$U$$ has a convergent subsequence that converges in the closure of $$U$$.
7. If in addition, $$U$$ is closed, then the subsequence will converge in $$U$$ itself.

For completeness of $$n$$-dim normed linear spaces and compactness of their closed and bounded subsets, check the notes here.