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Let $(V,T)$ be a topological vector space over $\mathbb{F}$ where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$.

Claim:

Let $U$ be an open neighbourhood of the origin. Then there is an open neighbourhood $N$ of the origin s.t $ \ \alpha N \subset U \ \forall \alpha \in \mathbb{F}$ s.t $|\alpha|\leq 1$.

Here is what I got.

Let $U$ be an open neighbourhood of the origin, form continuity, $U^{-1}$ is open in $\mathbb{F} \times V$ and contains $(0,0)$. Since $U^{-1}$ is open then it is contained in the product topology on $\mathbb{F} \times V$. Thus there is some sets $B_{r_{1}}, B_{r_{2}}, \dots$ and $T_1, T_2 \dots$ in the product topology s.t the union of those sets is equal to $U^{-1}$. Then for example we have that $B_{r_{1}} \subset U^{-1}$. My professor claims that $B_{r_{1}} \subset U^{-1} \Rightarrow \ \forall \alpha$ s.t $|\alpha| < r_1$ We have that $ r_1T_1 \subset U $. Can someone explain why we have this implication?

If this implication is true, then I can finish the rest.

Olba12
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  • what are "those sets" in "the union of those sets is equal to $U^{−1}$" ? You may have missed something from the explanation of your teacher. I would think that $B_{r_n}$ are open in $\Bbb F$, and the $T_n$ are open in $V$. Then $B_{r_n}\times T_n$ is open in $\Bbb F\times V$. – Mirko Dec 06 '16 at 02:52
  • They are element/sets of the basis, every set can be written as a union of the base elements. Yes $B_{r_n} \times T_n $ are open in $\mathbb{F} \times V$, what does this tell us? @Mirko – Olba12 Dec 06 '16 at 09:01
  • What do you mean by $U^{-1}$? – Moishe Kohan Dec 06 '16 at 10:01
  • Preimage @MoisheCohen – Olba12 Dec 06 '16 at 11:34
  • @Olba12: Preimage under what map? – Moishe Kohan Dec 06 '16 at 14:26
  • The map from FxV to V @MoisheCohen – Olba12 Dec 06 '16 at 14:27
  • I see. The implication you wrote is incorrect, it is $\forall \alpha, |\alpha|<r_1 \Rightarrow \alpha T_1\subset U$. This is indeed true and immediate from the inclusion ${\alpha} \times T_1\subset B_{r_1} \times T_1 \subset U^{-1}$. – Moishe Kohan Dec 06 '16 at 15:26
  • Ah yes I did Intent to write $\forall \alpha, |\alpha | < r_1 \Rightarrow \alpha T_1 \subset U$ sloppy of me. Now I see it. Thanks @MoisheCohen – Olba12 Dec 06 '16 at 15:34

1 Answers1

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Here is what your question is missing:

  1. Per your comment: $U^{-1}$ is the preimage in $F\times V$ of $U$ under the product map $F\times V\to V$.

  2. $B_{r}=\{\alpha\in F: |\alpha|< r\}$.

  3. $r_1$ is defined as a positive real number such that $(0,0)\in B_{r_1}\times T_1\subset U^{-1}$. This number exists by the definition of the standard basis of the product topology and the fact that $(0,0)\in U^{-1}$.

With these missing pieces in mind, the proof is immediate:

You have $B_{r_1} \times T_1 \subset U^{-1}$; since the product map sends $U^{-1}$ into $U$, it also sends $B_{r_1} \times T_1$ into $U$.

Moishe Kohan
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