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  1. Prove that a closed subspace of a Banach space is also a Banach space.
  2. Show that the linear space of all polynomials in one variable is not a Banach space in any norm.
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Welcome to math.SE! What have you tried so far? For example, for the first problem, you have to show that a closed subspace is complete with respect to the norm inherited from the norm on the initial Banach. But in fact it doesn't have anything to do with vector space (it's true that a closed part of a complete metric space is complete). For the second exercise, you can use Baire's categories theorem. – Davide Giraudo Apr 10 '12 at 18:51
Is it just me or does one of these problems seem substantially more difficult than the other? – David Mitra Apr 10 '12 at 19:23
@DavidMitra: it's not just you :) – t.b. Apr 10 '12 at 19:26


  1. Prove that a closed subspace of a complete metric space is complete.

  2. The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire category theorem.

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The argument in the second exercise is easily adapted to the statement that the vector space dimension of a Banach space is either finite or uncountable. With different methods one can establish that the dimension of an infinite-dimensional Banach space is at least the cardinality of the continuum. – t.b. Apr 10 '12 at 19:28
Because a subspace with non-empty interior is open and an open subspace of a topological vector space is the entire space, for example by connectedness (an open subspace is closed). – t.b. Jul 2 '12 at 13:35
I think I do answer your question: I argue that a subspace with non-empty interior has to be the entire space and the space polynomials of degree $\leq n$ (or more generally a proper subspace, closed or not) is not the entire space. – t.b. Jul 2 '12 at 13:59
No worries :) Maybe you prefer this argument: if a subspace has non-empty interior it contains some ball $B_\varepsilon(x)$ with $x$ in the subspace. Translating by $-x$ we see that $B_\varepsilon(0)$ is contained in the subspace. By homogeneity of the norm and closedness of the the subspace under scalar multiplication the subspace has to be the entire space. – t.b. Jul 2 '12 at 14:11
@MattN. Let $Y$ be the subspace of $X$ and let $y \in Y$ be an interior point. Then there is an open set $U$ of $X$ such that $y \in U \subset Y$. For every $y' \in Y$ we have that $y' - y + U$ is an open neighborhood of $y'$ in $X$ and contained in $Y$, so $Y$ is open. – t.b. Jul 29 '12 at 19:18

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