Prove: sum of a finite dim. subspace with a subspace is closed

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SUMMARY

The discussion centers on the proof that the sum of a finite-dimensional subspace \( L \) and any subspace \( M \) of a locally convex space \( X \) is closed. It is established that if \( \dim L < \infty \), then \( L \) is closed in \( X \). The conclusion drawn is that \( L + M \) is closed in \( X \) without the necessity for \( M \) to be closed or finite-dimensional.

PREREQUISITES
  • Understanding of locally convex spaces
  • Knowledge of finite-dimensional subspaces
  • Familiarity with the properties of closed sets in topological spaces
  • Basic concepts of vector space operations
NEXT STEPS
  • Study the properties of locally convex spaces in detail
  • Learn about the closure of subspaces in topological vector spaces
  • Explore the implications of finite-dimensionality on vector space operations
  • Investigate examples of closed and non-closed subspaces in various vector spaces
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Mathematicians, particularly those specializing in functional analysis, students studying topology, and anyone interested in the properties of vector spaces and their subspaces.

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Homework Statement


Prove:
If ##X## is a (possibly infinite dimensional) locally convex space, ##L \leq X##, ##dimL < \infty ##, and ##M \leq X ## then ##L + M## is closed.

Homework Equations


The Attempt at a Solution



##dimL < \infty \implies L## is closed in ##X##
##L+M = \{ x+y : x\in L, y \in M \} \implies ^{??} dim(L+M) < \infty \implies L+M ## is closed in ##X##

Homework Statement


Homework Equations


The Attempt at a Solution

 
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Don't you need M to be closed as well??

Anyway, your attempt isn't correct since L+M doesn't need to be finite-dimensional.
 

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