I need some help with the proof I have made. This post is closely related to the attached post but this is a proof verification with some additional questions.
We are given an inner product space $X$ with a sequence $x_n$. We assume that $x_n\rightarrow x$ and $x_n \perp y$, and we want to show that $x\perp y$.
My take:
Since $x_n\perp y$ we know that $(x_n,y)=0$ where $(\;\cdot\;,\;\cdot\;)$ is our inner product. We can write $$ |(x_n,y)-(x,y)|=|(x_n-x,y)|\leq \|x_n-x \|\| y\| $$
We have now a bound on $(x_n,y)-(x,y)$ and since $x_n\rightarrow x$ we can make $\| x_n-x \|$ arbitrarily small, thus also making $(x_n,y)-(x,y)$ arbitrarily small which implies $(x_n,y)=0=(x,y)$ and finally $x\perp y$.
My question is now if this proof is correct. Also, I feel a little uncertain about convergence in inner product spaces. When we work in inner product spaces and say that a sequence convergence (without any additional information), do we mean the "normal" epsilon-delta convergence? Is this the same as strong convergence? This is my first post here, feel free to add comments about how to improve my way of asking questions.
$x_n \to x$ in an inner product space if the sequence of numbers $\|x_n-x\| \to 0$. What you have done is OK. To give a precise argument you can separate the cases $y=0$ and $y \neq 0$. When $y \neq 0$, given $\epsilon >0$ there exist $n_0$ such that $\|x_n-x\| <\epsilon /||y\|$ for $n >n_0$. This gives $|(x_n,y)-(x,y)| <\epsilon$ for $n >n_0$. This implies $|(x,y)| <\epsilon$. Since $\epsilon $ is arbitrary this gives $(x,y)=0$.