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.

  • 7
    $\begingroup$ 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 $\mathbb{R}^n$ $\endgroup$
    – matgaio
    May 13, 2012 at 4:01

2 Answers 2


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


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


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 $\{v_k\}_{k=1}^\infty$.

  • $\begingroup$ 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. $\endgroup$
    – Our
    Nov 4, 2017 at 10:27
  • 1
    $\begingroup$ @Our Norms on finite dimensional vector space are equivalent. $\endgroup$
    – 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.


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