Prove that if $a\in\text{int}(S)$ and $b\in\text{cl}(S)$ then $[a,b)\subseteq\text{int}(S)$ when $S$ is convex. Definition
Let be $X$ a topological vector space. A subset $S$ of $X$ is said convex if the affine combination
$$
A:=\{z\in X: z=(1-t)x+ty, t\in[0,1]\}
$$
is contained in $S$ for any $x, y\in S$.
Statement
If $[a,b):=\{(1-t)a+tb:t\in[0,1)\}$ then $[a,b)\in\text{int}(S)$ when $a\in\text{int}(S)$, when $b\in\text{cl}(S)$ and when $S$ is convex.
Clearly $[a,b)\subseteq\text{cl}(S)$ because $S$ is convex and so $\text{cl}(S)$ too but how to prove that actually $[a,b)\in\text{int}(S)$? So could someone help me, please?
 A: Let $(b_n)\in S^\Bbb{N}$ be a sequence of points in $S$ that converges to $b$ and let $r>0$ be such that $\overset\circ{B}(a,r)\subset S$. Let $c\in[a,b)$: there exists $t\in(0,1]$ with $c=ta+(1-t)b$ and let us put, for $n\in\Bbb{N}$, $c_n=ta+(1-t)b_n$. It is easy to show that for all $n\in\Bbb{N}$, $\overset\circ{B}(c_n,tr)\subset S$. Now
$$\forall n\in\Bbb{N},~\|c-c_n\|=(1-t)\|b-b_n\|$$
Since $b_n\to b$ and $tr>0$, there exists $N\in\Bbb{N}$ such that $(1-t)\|b-b_N\|<tr$ and thus $c\in\overset\circ{B}(c_N,tr)\subset \overset\circ{S}$.
EDIT. For TVS. Let $\mathcal{V}$ be the directed set of open neighborhoods of $0$. For every $V\in\mathcal{V}$ there exists $b_V\in (b+V)\cap S$ and we put $c_V=ta+(1-t)b_V$. Let $\mathcal{O}\in\mathcal{V}$ be a symmetric ($\mathcal{O}=-\mathcal{O}$) open neighborhood of $0$ such that $a + \mathcal{O}+\mathcal{O}\subset S$. Note that for all $V\in\mathcal{V}$, $c_V+t\mathcal{O}+t\mathcal{O}\subset \mathrm{int}(S)$. Furthermore, for any $x\in X$,
$$
x\in c+t\mathcal{O}
\implies
c+t\mathcal{O}  \subseteq x +t\mathcal{O}+t\mathcal{O}.
$$
Since the net $(b_V)_{V\in\mathcal{V}}$ converges to $b$, the net $(c_V)_{V\in\mathcal{V}}$ converges to $c$ and there exists an open neighborhood $V_0$ of $0$ such that $c_{V_0}\in c+t\mathcal{O}$ and therefore $c+t\mathcal{O} \subseteq c_{V_0} +t\mathcal{O}+t\mathcal{O}\subset \mathrm{int}(S)$; in particular $c\in\mathrm{int}(S)$.
