$\lim_{x\to 0}\sqrt{x}=0$ reasoning/relation to $\epsilon$-$\delta$ definition The following definition below is what I believe is the definition of a limit. However, there might be something subtle that I am missing.
Definition of a Limit: Let $f: X\to Y$ be defined on some open interval $I\subseteq X$ that contains $a$, except possibly at $a$ itself. Then,  $\lim_{x\rightarrow a}f(x):=L$  if $\forall \epsilon \in \mathbb{R}^{>0}$, $\exists \delta\in \mathbb{R}^{>0}$, $\forall x\in I [\text{ if } 0 <|x-a|< \delta \text{, then } |f(x)-L|<\epsilon].$
Now, consider $f(x):=\sqrt{x}$ where $f$ is a function such that $f: [0, \infty)\rightarrow \mathbb{R}$. Why would $\lim_{x\to 0}f(x)=\lim_{x\to 0}\sqrt{x}=0$? Wouldn't it not make sense to figure out the limit here because we can't find some open interval $I$ that satisfies this? I'm looking for clarification on this definition to ensure this scenario makes sense. Any help would be greatly appreciated!
Definition: Let $A\subseteq X$ where $X$ is a metric space. Let $x\in X$. We say $x$ is a limit point of $A$ if $\forall B_x$ (i.e. an open ball with center x) $(B_x-\{x\})\cap A\neq \emptyset$.
 A: Note that $X$ is a space that does not have to be equal to $\mathbb R$.
The definition of an interval in the space $X$ is any set $I$ so that for any $a, b \in I$ then the all the elements $w\in X$ so that $\min(a,b) \le w \le \max(a,b)$ will be $w\in I$.
So although the $S=(1,2)\cup \{3\}\cup (4, 5)$ will not be an inteval in $\mathbb R$,  $S$ will be an interval in the space $X= (-\infty, 2)\cup \{3\} \cup (4,\infty)$.
The definition of an open interval is an interval $I$ so that for every $a \in I$ there exists a $\delta > 0$ so that $x\in S$ so that $|x-a| < \delta$ will be that $x\in I$.
By that definition $[0, 1)$ is not an open interval in $\mathbb R$.  It fails because $0\in I$ but for any $\delta> 0$ we can find an $x: -\delta < x < 0$ and $|x-0| < \delta$ but $x \in [0,1)$ so $[0,1)$ can not be an open interval in $\mathbb R$.
But $[0,1)$ is an open interval in $[0,\infty)$.  Why?  Consider $0 \in [0,1)$.  Let $\delta = \frac 12$.   If $|x- 0| < \frac 12$ and $x\in [0,\infty)$ then $0\le x < \frac 12$.  And so $x \in [0,\infty)$.  So $0$ being an endpoint of $[0, 1)$ with nothing to the left of it does not cause $[0,1)$ to fail to be an open interval.  It's okay that there is nothing to the left of the point $0$ in $[0,1)$ because..... there is nothing to left of $0$ in our entire universe space of $[0, \infty)$.
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So according to you definition $\lim_{x\to 0} f(x) = 0$ where $f:[0,\infty)\to \mathbb R:  f: x\to \sqrt x$ works out just fine.
"Let $f:[0,\infty)→\mathbb R$ be defined on some open interval $I\subset [0,\infty)$ that contains $0$ except possibly at a itself".
Okay... $[0,1)$ is such an open interval.  $[0,1)$ is open in $[0,1)$ and $\sqrt x$ is defined on $[0,1)$.
"The $\lim_{x\to 0} f(x) = 0$ if $\forall \epsilon > 0$ there exists a $\delta >0$"
Okay.... let $\delta = \epsilon^2$.
"So that for all $x\in [0, 1)$ if $0< |x-0| < \delta$ then $|f(x) -0| < \epsilon$".
That follows.
If $x \in [0,1)$ and of $0 < x-0 < \delta$ then $-\delta < x < \delta$ and as $0 \le x \le 1$ we have $0 \le x < \delta$.
And as $\delta =\epsilon^2$ we have $0 \le x < \epsilon^2$ so $0 \le \sqrt x < \epsilon$.
So $0 \le f(x) < \epsilon$ so $|f(x) -0| = f(x) < \epsilon$.
The definition is satisfied.
