Topological interior point is algebraic interior point in general TVS

Question

Let $X$ be a topological vector space and let $S \subseteq X$ a subset.

A point $y \in S$ is called algebraic interior point if for every $x \in X$ there exists a $\varepsilon_{x} > 0$ s.t. $\forall t \in (-\varepsilon_{x}, \varepsilon_{x})$ we have $y + tx \in S$.

Now I have a note saying in a topological vector space any topological interior point is an algebraic interior point. If the space is normed, this is the case:

Let $y \in \text{int}(S)$, then since $\text{int}(S)$ is open, there exists a $\varepsilon > 0$ s.t. $B_{\varepsilon}(y) \subseteq \text{int}(S) \subseteq S$. So let $x \in X$ arbitrary, then $\exists \alpha > 0$ s.t. $ \Vert y - y + \alpha x \Vert = \vert \alpha \vert \Vert x \Vert < \varepsilon$ and since the open ball is convex, the line segment is contained in it.

However, I am not able to get a proof for a general topological vector space. Also I am having the following doubt:

Let $X$ be a real vector space of dimension $n >1$ and let its topology be generated by a one dimensional subspace i.e. let $V \subset X$ with $\dim(V) = 1$ be open. Then every $y \in V$ is a topological interior point since $V$ is open, but none of them is an algebraic interior point. For this take any $x \not\in V$, then $\alpha x \not\in V$ for $\alpha \neq 0$ and thus $y + \alpha x \not\in V$.

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Actually, the original claim does not even seem to be true for a metric vector space. Consider any real vector space with the discrete metric and consider a singleton $\{y\}$. Then $y$ is a topological interior point, since the singleton is open. However, for any other $x \in X$ we have $y+tx \not\in \{y\}$ for $t \neq 0$.

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Is my note wrong? If not, I am looking for a proof that for a general topological vector space, a topological interior point is an algebraic interior point.

Answer 1

The answer lies is the definition of a topological vector space. We ask that the maps $\mathbb{R}\times X \rightarrow X$ and $X\times X \rightarrow X$ given by multiplication and addition are continuous :

Let X be a topological vector space, $S\subset X$, and $y\in int(S)$. We can suppose that $S$ is open. The map $\phi:\mathbb{R} \rightarrow X$, given by $t\mapsto y+tx$ is continuous, and $\phi(0) = y\in S$, so by continuity, there exists $\epsilon_x$ such that $\phi(]-\epsilon_x,\epsilon_x[) \subset S$.

Answer 2

Your question was already nicely answered by Maxime. I will adress your examples.

The main point is that a topological vector space is not just any vector space with a topology, but a vector space with a topology in which linear operations are continuous.

Note that a vector space with dimension at least one with a discrete metric is NOT a topological vector space as multiplication is not continuous. To see that consider the multiplication operator $$M: \mathbb{R} \times X \rightarrow X; \;\;\; M(\alpha,x) = \alpha x.$$ Then for $x \neq 0$ we have $(\frac{1}{n},x) \rightarrow (0,x)$ in $\mathbb{R} \times X$ but $M(\frac{1}{n},x) = \frac{1}{n} x \not \rightarrow 0$ in $X$ as only eventually constant sequences are convergent in the discrete topology.

Similarly note that no proper vector subspace $V$ of a topological vector space $X$ can be open as its complement is not closed. To see that consider any $x \notin V$. Then $\frac{1}{n} x = M(\frac{1}{n},x) \rightarrow M(0,x) = 0 \in V$.