# Generalization topological vector space

A subset $A$ of a topological space $X$ is said to be semi-open if there exists an open set $O$ in $X$ such that $O \subset A \subset Cl(O)$, or equivalently if $A \subset Cl(Int(A))$. $SO(X)$ denotes the collection of all semi-open sets in $X$.

Recall that a set $U \subset X$ is a semi-open neighborhood of a point $x \in X$ if there exists $A \in SO(X)$ such that $x \in A \subset U$. A set $A \subset X$ is semi-open in $X$ if and only if $A$ is semi-open neighborhood of each of its points. If a semi-open neighborhood $U$ of a point $x$ is a semi-open set, we say that $U$ is a semi-open neighborhood of $x$.

Definition An s-topological vector space $(X , \tau )$ is a vector space $X$ over the field $\mathbb F$ ($\mathbb R$ or $\mathbb C$) with a topology $\tau$ defined on $X$ and standard topology on $\mathbb F$ such that:

1) For each $x, y \in X$, and for each open neighborhood $W$ of $x + y$ in $X$, there exist semi-open neighborhoods $U$ and $V$ of $x$ and $y$ respectively in $X$, such that $U + V \subseteq W$.

2) For each $\lambda \in \mathbb F$, $x \in X$ and for each open neighborhood $W$ of $\lambda x$ in $X$, there exist semi-open neighborhoods $U$ of $\lambda$ in $\mathbb F$ and $V$ of $x$ in $X$ such that $UV\subseteq W$.

I need to prove that the following example is an s-topological vector space

Example: Let $X = \mathbb R$ be a vector space of real numbers over the field $\mathbb F = \mathbb R$ and let $\tau$ be a topology on $X$ induced by open intervals $(a, b)$ and the sets $[1, c)$ where $a, b, c \in\mathbb R$.

You can say in your example that:

(i) $\overline {[1,c)} =[1,c]$

(ii) Every $U \in \tau$ is of the form $A \cup [1,c)$, while $A$ is open in the usual topology.

Also for every $U,V \in \tau$, you get that $\overline {U \cup V} =\overline U \cup \overline V$, so $\overline {A \cup [1,c)}=\overline A \cup [1,c]$.

From your defenition it's also clear that every open set is also semi open, and you can also note that the intervals $[a,b),(a,b]$ are also semi open sets.

1) If $x+y \neq 1$ then there exists $\delta >0$ such that $(x+y) \in(x+y-\delta, x+y-\delta)$ and $(x+y-\delta, x+y-\delta) \subseteq W$. So the semi-open neighbourhoods $U=(x-\frac{\delta}{2},x+\frac{\delta}{2})$ and $V=(y-\frac{\delta}{2},y+\frac{\delta}{2})$ fulfill requirement (1).

If $x+y=1$ then there exists $\delta >0$ such that $x+y \in [1,1+\delta) \subseteq W$. Then the semi-open neighbourhoods $U=[x,x+\frac{\delta}{2})$ and $V=[y,y+\frac{\delta}{2})$ fulfill requirement (1).

2)If $\lambda x \neq1$, then there exists $\delta >0$ such that $\lambda x \in (\lambda x -\delta ,\lambda x +\delta) \subset W$. Then there exists $\delta_1,\delta_2>0$ such that $|\lambda \delta_2 +x \delta_1 +\delta_1 \delta_2| <\delta$. Then the semi open neighbourhoods $U=(\lambda -\delta_1, \lambda +\delta_1)$ and $V=(x- \delta_2,x+ \delta_2)$ fulfill requirement (2).

If $\lambda x=1$, then there exists $\delta >0$ such that $\lambda x \in [1 ,\lambda x +\delta) \subset W$. Then there exists $\delta_1,\delta_2>0$ such that $|\lambda \delta_2 +x \delta_1 +\delta_1 \delta_2| <\delta$. Then the semi open neighbourhoods $U=[\lambda , \lambda +\delta_1)$ and $V=[x,x+ \delta_2)$ fulfill requirement (2).

• thank you very much for your answer by the way the semi-open sets U and V that you used it it is also open is that true – x and y Mar 18 '17 at 12:10
• I'm sorry but I don't understand your question. – Keen-ameteur Mar 18 '17 at 12:14
• I'm sorry also because i'm not explain my question very well here is my question: in case 1 and two you use the semi-open neighborhoods U and V – x and y Mar 18 '17 at 18:22
• I'm sorry also because i'm not explain my question very well here is my question:In case one and two you use the semi-open neighborhoods U and V but i think that these two sets U and V are also open neighbourhoods – x and y Mar 18 '17 at 18:28
• If I understand your question correctly When $\lambda x \neq1$ or $x+y \neq 1$, I do use open neighbourhoods (which are also semi open) because it's the simplest argument in my mind. But in the case of $\lambda x =1$ or $x+y=1$ you have to use semi-open neighbourhoods which are not open. – Keen-ameteur Mar 19 '17 at 6:33