Consider a topological vector space $(X,\tau_X)$ and $N_0$ a neighborhood of $0$. I want to show, that it is absorbent, i.e. $X=\cup_{t> 0} tN_0$.
My idea would be to take an arbitrary $x_0\in X$ and to find a $\delta > 0$, such that $\delta x_0\in \operatorname{int} N_0$. Then, you could see, that $x_0\in\delta^{-1} N_0$.
However, I do not exactly know, why there is this $\delta$, such that $\delta x_0\in N_0$. My impression is, that otherwise $0$ must be a boundary point of $N_0$ ? Which would be a contradiction to the fact, that $N_0$ is a neighborhood of $0$?
I know, that the scalarmultiplication $s_\delta:X\rightarrow X$, $s_\delta(y):=\delta y$ is continuous, but I am not sure how to use this fact here, because of the following:
By definition, a function $f$ is continuous in point $x$, if for any $N$, a neighborhood of $f(x)$, there is a $M$, a neighborhood of $x$, such that $f(M)\subseteq N$.
To use this definition, wouldn't I need to use the scalar multiplication $s_0:X\rightarrow X$, $s_0(y)=0$? (For mapping a neighborhood of $x_0$ into $N_0$, a neighborhood of $0=s_0(x_0)$?)
Any help or hint is appreciated! Thank you in advance!
By definition of a TVS, the map $$m:\mathbb{R}\times X\rightarrow X,\quad (t,x)\mapsto tx$$ is continuous. Pick $x_0\in X$. For an open set $U\subseteq X$, which contains $0$, the set $m^{-1}(U)\subseteq \mathbb{R}\times X$ is open and contains $(0,x_0)$. Since $$\mathcal{B}_{\mathbb{R}\times X}=\{B_1\times B_2:B_1\in\mathcal{B}_{\mathbb{R}}, B_2\in\mathcal{B}_{X}\}$$ is a basis of the topology on $\mathbb{R}\times X$, there must be $\epsilon >0$ and an open $B_X\subseteq X$, such that $$(0,x_0)\in(-\epsilon,\epsilon)\times B_X\subseteq m^{-1}(U).$$ From this, one finds, that for any $\delta\in(-\epsilon,\epsilon)\backslash\{0\}$, it holds $\delta x_0=m(\delta,x_0)\in U$, such that $$x_0\in \delta^{-1}U\subseteq\cup_{t>0}tU.$$