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Let $X$ be a topological vector space and $x^\star, y^\star\in X$. Let $\lambda \in (0, 1)$. We define a map $f: X \times X \to X$ by $f(x, y) = \lambda x + (1-\lambda)y$. Then $f$ is continuous. Let $z^\star := f(x^\star, y^\star)$ and $U$ be an open neighborhood of $z^\star$. Then there is an open neighborhood $V$ of $(x^\star, y^\star)$ such that $f(V) \subseteq U$.

In this answer, @Seirios goes further and asserts that there are some open neighborhood $V_1$ of $x^\star$ and open neighborhood $V_2$ of $y^\star$ such that $f(V_1 \times V_2) \subseteq U$.

Could you elaborate how to obtain such $V_1$ and $V_2$?

Akira
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1 Answers1

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If $V$ is open, it is (by definition) the union of basic open neighborhoods of the form $V_1 \times V_2$. As $V$ contains $(x^*, y^*)$, one of these basic neighborhoods must contain $(x^*, y^*)$.

Joseph Camacho
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  • hi what is a basic open neighborhood? is it an element of a neighborhood basis? How does that concecpt relate to the basis of a topology? Because I only know that an open set is a union of basic sets not of basic neighborhoods. – guest1 Apr 04 '23 at 11:49