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While studying some basic measure theory, I stumbled upon some really basic questions I couldn't really find a simple explanation or proof for.

Is it true for real intervalls and a convergent sequence $c_i$ that $$\bigcup_{i=1}^{\infty}\,[a+c_i, b] = \lim_{i\to \infty} [a+c_i,b]\quad?$$ The reason why I am asking that is because I am well aware that, for example, $$\bigcup_{i=1}^{\infty}\,\left[a, b - \frac{1}{i}\right] = [a, b[ $$ but since $\frac{1}{i} \to 0$ I struggle with understanding why it is not equal to $[a_i, b_i + 0] = [a_i. b_i].$ I do intuitively why it is not the closed interval, I just want my definitions to be consistent. I'd appreciate an explanation.

Edit: Thanks for your comments/answers so far. My big problem is understanding the difference of taking limits and taking countable unions and/or intersections I think. For example, if you look at the sequence $1/n$, we have that $0 \notin\{1/n\}$ for all $n\in \mathbb N$ but still $\lim_{n\to \infty} 1/n = 0.$ So the argument $$\bigcup_{i=1}^{\infty}\,[a,b-1/i] \neq[a,b]$$ since $b\notin [a,b-1/i]$ for all $i\in \mathbb N$ isn't really intuitive for me because we are looking at some kind of "limit". I hope I now made it more clear.

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    Did you mean an intersection in the last equation? – 2017-02-28
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    Oh no, I meant to put a minus-sign. – 2017-02-28
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    The last equation is incorrect as @Qudit rightly notes – 2017-02-28
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    I corrected it. – 2017-02-28
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    It's still incorrect even with the change because the $i$ is not bound. – 2017-02-28
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    I edited and made my question more precise. – 2017-02-28
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    It is certainly true for _some_ families of real intervals that $\bigcup_{i=1}^{\infty}\,[a_i, b_i] = \lim_{i\to \infty} [a_i,b_i].$ Just set $a_i = m < n = b_i$ for all $i,$ where $m$ and $n$ are constants. You can prefix any number of subintervals of $[m,n]$ to the list and get the same result. But this is an unusual property for a sequence of intervals or sets in general, even if the sequence converges. [This comment possibly rendered obsolete by an edit to the question.] – 2017-02-28
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    I can easily think of multiple ways to make $\bigcup_{i=1}^{\infty}\,[a+c_i, b] \neq \lim_{i\to \infty} [a+c_i,b].$ Setting $c_i = 1/i$ will do it; so will defining any negative value of $c_i.$ It might be helpful to know which "definitions" you want to be consistent. (None of the equations you've shown is a definition.) – 2017-02-28
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    Actually, I wonder what's the difference in the definition of a finite union and a countable (or even uncountable) union. I always thought of it as a limit. – 2017-02-28
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    It seems from the edit that we need to notice the limit of that sequence approaches the value, but never quite attains it? Then we see that the interval $[a,b-\frac{1}{i}]$ approaches $[a,b]$, but since we never got to $b$ we have $[a,b)$... – 2017-03-01

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$b$ is clearly not in the union as it is not a member of any $[a, b-\frac{1}{i}]$.

But if $a< x < b$, $x - b > 0$ and so for some $i$ $0 < \frac{1}{i} < (b-x)$. This implies that $x \in [a, b-\frac{1}{i}]$ so in the union on the left. Hence the $[a,b)$ result.

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    I updated my question and made it more precise. My main question was if that relation between taking countable intersections/unions and taking limits holds and if so how to think about it in a correct way. – 2017-02-28
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    Limits of sets are unions or intersections. You cannot interchange the limits of interval boundaries with the limits of sets just like that. You have to think about it every time. – 2017-02-28
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    Thanks. That is probably the thing I am struggling with right now. If you look at my last edit, I explained a little bit more what my problem is. :) – 2017-02-28