$x\in U\implies x\in (1-\delta)U$

Question

Let $X$ be a topological $\mathbb{R}$-vector space. I want to show that if $U$ is open and $x\in U$ then there exists a $\delta>0$ such that $x\in (1-\delta)U$. I think it has something to do with the continuity of the scalar multiplication $\mathbb{R}\times X\to X$. Since $x=1\cdot x$ I know that by the continuity of this scalar multiplication that there exists a neighborhood $V$ of $x$ and $(1-\delta,1+\delta)$ of $1$ such that $(1-\delta,1+\delta)V\subset U$ but I am not sure how to continue.

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By the above reasoning, there exists $\epsilon>0$ such that $(1-\epsilon,1+\epsilon)V\subset U$, so there exists $\eta>1$ such that $\eta x\in U$ then take $\delta:=\frac{\eta-1}{\eta}>0$, then $\frac{1}{1-\delta}x=\eta x\in U$ so $x\in (1-\delta)U$.

Answer 1

$(1+\frac 1 n )x \to (1)(x)=x$. So there exists $n_0$ such that $(1+\frac 1 {n_0} )x \in U$. This means $x \in (1-\delta )U$ where $\delta =\frac 1 {n_0+1}$.

Answer 2

If scalar multiplication is denoted by $s: \Bbb R \times X \to X$ we see that $s((1,x)=1 \in U$ so there exists a basic product neighbourhood $(1-\delta,1+\delta) \times V$ of $(1,x)$ so that $s[(1-\delta,1+\delta) \times V ] \subseteq U$ by pointwise continuity. Take that $V$ and $\delta$.