I was watching a youtube tutorial about pointwise convergence and the author was interpreting the following piecewise function:
$$h_{n}(x) = \left\{ \begin{array}{ll} 1,& x\geq \frac{1}{n} \\ nx & 0 \leq x < \frac{1}{n}\\ \end{array} \right. $$
Specifically there is one thing that he said that concerned me about how I'm interpreting these things. Considering the case where $ 0 < x < \frac{1}{n}$. I purposely omitted the endpoint $0$ as I'm just trying to gather what is occuring beyond there. So the author said that if we take $\lim_{n\rightarrow \infty}$ then $\frac{1}{n} \rightarrow 0$. I agree with this. But then next he said that our value of $x$ which originally is in $ 0 < x < \frac{1}{n}$ get's moved into the other interval $x\geq \frac{1}{n}$. Is this true?....I was always of the idea that if $\lim_{n\rightarrow \infty}$ and $x$ was in $ 0 < x < \frac{1}{n}$ that $x \rightarrow 0$ because the interval is shrinking and $x$ is in it. Visually if this was on the number line I pictured $x$ getting pushed to the left towards $0$. So have I been interpreting this wrong and actually $x$ remains fixed where it is and instead moves into the other interval?, specifically $x\geq \frac{1}{n}$.
In your piecewise function definition, $x$ and $n$ are shown as variables which are independent of each other. For a given value of $x$ and $n$, the value of $h_n(x)$ depends on which case $x$ falls into based on the value of $n$. When you state
At this point, neither $x$ nor $n$ values are specifically stated, but you state the relation above holds for these particular values.
Now, if with the value of $x$, whatever it is, you next try to determine $\lim_{n\to \infty}h_n(x)$, note that only $n$ is changing, and it's original value doesn't matter. As long as $x \gt 0$, at some point as $n \to \infty$, you'll get $n \gt \frac{1}{x}$, so $x \ge \frac{1}{n}$ and $h_n(x) = 1$, giving that $\lim_{n\to \infty}h_n(x) = 1$ for $x \gt 0$. Of course, in the original set of cases, if $x = 0$, then $h_n(x) = 0$ for all $n$, so $\lim_{n\to \infty}h_n(0) = 0$.
What you're asking about happening, i.e., the $0 \lt x \lt \frac{1}{n}$ condition remains, could be indicated by something like $\lim_{n \to \infty, 0 \lt x \lt \frac{1}{n}}h_n(x)$ instead. This then forces this extra constraint to always hold as $n \to \infty$. However, the limit would not exist in this case as the value of $h_n(x)$ would be $nx$ while $0 \lt nx \lt 1$ so there's no specific values it's going to unless there is another more stringent constraint on how $x$ changes as $n \to \infty$.