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Given a continuous linear map $T:E\to F$ where $E,F$ are normed vector spaces, I am wondering about the trivial question whether for any subset $U\subset E$, it holds or not: $$T(U^c)=T(U)^c$$ I could make sense of this just under the assumption that $T$ is surjective, but I am not sure if the following reasoning is correct:

  • trivially we have that $E=U\oplus U^c$, therefore $T(E)=T(U)\oplus T(U^c)$ by linearity of $T$.
  • by surjectivity we have that $T(E)=F$, hence $F=T(U)\oplus T(U^c)$ $\underline{which \ implies}$: $$T(U^c)=T(U)^c\ $$

I am not really sure about the underlined implication. Any idea?

mastro
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This is false: since $T(U)$ for any subspace $U$ is itself a linear subspace, it contains the zero element. This implies that the complement of a subspace cannot contain zero.

daw
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  • what you said is wrong. The complement of a vector space is a vector space, and indeed it must contains the zero-element. In fact, given a vector space $X$ one has that: $$X\cap X^c={0}$$ – mastro Jan 16 '15 at 09:35
  • @Filippo There is some confusion here between complement (normally denoted as $X^c$ and defined as the set of elements that do not belong to $X$) and orthogonal complement usually denoted as $X^{\bot}$. – drhab Jan 16 '15 at 09:53
  • @drhab Let $G ⊂ E$ be a closed subspace of a Banach space $E$. A subspace $L ⊂ E$ is said to be a topological complement or simply a complement of $G$ if:
    • (i) $L$ is closed,
    • (ii) $G∩L={0}$ and $G+L=E$.
     – mastro Jan 16 '15 at 09:59
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    then please clarify your question: I have never seen the notation $U^c$ to be used for complement subspace. Moreover, the notation suggests that the complement is uniquely defined. – daw Jan 16 '15 at 10:26