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Let $E_1, E_2, \ldots, E_n, \ldots$ be linear spaces over the same field E. Denote $U = \bigcup_{k=1}^{\infty} E_k$ and $I = \bigcap_{k=1}^{\infty} E_k$.

Answer: $\bigcup_{k=1}^{\infty} E_k$: The union $U$ is a linear space because each $E_k$ is a linear space, and the union of linear spaces is also a linear space. This corresponds to the condition "The union of linear spaces is a linear space."

$\bigcap_{k=1}^{\infty} E_k$: The intersection $I$ is also a linear space. To show this, we need to verify whether $I$ is closed under addition and scalar multiplication. Suppose $v, w \in I$, meaning that $v, w \in E_k$ for all $k$. If $v, w \in I$, then $v - w \in E_k$ for all $k$ because $E_k$ is a linear space.

Now, it is essential to check for closure. Let $v_n$ be a sequence in $I$, i.e., $v_n \in E_k$ for all $k$. If $v_n \to v$ (a limit vector) and $v \in E_k$ for all $k$, then $v \in I$ because $I$ is closed under the condition of closure.

Thus, the union $\bigcup_{k=1}^{\infty} E_k$ and the intersection $\bigcap_{k=1}^{\infty} E_k$ are linear spaces.

Is this correct? Thanks for any help!

Temoi
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ugjumb
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  • This will be true if the sequence of linear spaces is an increasing sequence with respect to set inclusion, that is to say, $E_1 \subseteq E_2 \subseteq E_3 \dots.$ Also, the closure under taking limits does not make sense for arbitrary vector spaces. It should be a normed linear space or topological vector space. – Dhawal Patil Jan 12 '24 at 11:53
  • if it is not too much to ask can you show me how to write it correctly? – ugjumb Jan 12 '24 at 11:59

1 Answers1

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No. It is not correct. The sentence

The union of linear spaces is a linear space.

is not true.

See for example here or here.

In fact, the question doesn't really make sense. If you meant to write "linear subspaces", then it is not correct. If you really meant to write "linear spaces", it is much, much worse. Indeed, the union of two vector spaces could have two zero vectors, and the intersection of two vector spaces can be empty, hence cannot be a vector space. However, the intersection of vector subspaces is indeed a subspace.

Finally, the problem has nothing to do with functional analysis.

Plop
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  • Ah okay, but the intersection part is correct, yes? – ugjumb Jan 12 '24 at 11:02
  • The question seemed strange to me too, that is why I am having such difficulty, but I double checked it and yes it says vector/linear spaces not subspaces – ugjumb Jan 12 '24 at 11:25