Example that the union of an infinite number of closed sets is not necessarily closed.

This boils down to

1. "Infinity is not a number" and

2. $\bigcup\limits_{i=1}^\infty$ (or $\sum\limits_{i=1}^\infty$ or any index with $\infty$) is "just notation" and only means and means exactly what we say it means.

We use the notation $\bigcup\limits_{i=1}^{\color{red}n} A_i$ to be short-hand for $A_1\cup A_2\cup ..... \cup A_{{\color{red}n}-1} \cup A_{\color{red}n}$. That is notation and it means that because that is what we say it means.

That does not mean if I replaced $\color{red}n$ with something else that didn't make any sense it would mean the same thing. $\bigcup\limits_{i=1}^{-5}A_i$ can't mean $A_1 \cup A_2\cup .... \cup A_{-6} \cup A_{-5}$ because numbers just don't work that way and you can't count up to $-5$ starting at $1$.

And $\bigcup\limits_{i=1}^{\text{Big Bird, from Sesame Street}} A_i$ can not mean $A_1\cup A_2\cup ..... \cup A_{\text{Big Bird, from Sesame Street}-1} \cup A_{\text{Big Bird, from Sesame Street}}$ because that's just nonsense.

And likewise $\bigcup\limits_{i=1}^{\infty} A_i$ does not mean $A_1\cup A_2 \cup .... \cup A_{\infty - 1} \cup A_{\infty}$. That doesn't make sense because $\infty$ is not a number you count up to.

So what does $\bigcup\limits_{i=1}^\infty A_i$ actually mean?

Well we can have an union of a finite number sets... and we can have an union of an infinite number of sets. $A_1\cup A_2 \cup A_3 \cup ....... \cup A_{535798} \cup A_{535799} \cup ..... \cup A_n \cup A_{n+1} \cup ........ _{\text{it never ends}}...$

What should we use for notation for that?

Well, some bright guy had the idea that was both incredibly smart and incredibly stupid at the same time that if the notation of $\bigcup\limits_{i=1}^n A_i$ means the union of the the sets starting at $1$ and going up to $n$ then to to an infinite union we starting at one and never ending and we sometimes say "and go on to infinity"

(although if you think about it, that phrase "go on to infinity", although common is actually wrong-- we can't "go on" to infinity-- infinity is not a place we can "get to"-- we just... "go on" ... forever without end)

... and we sometimes say "and go on to infinity", then we can use the infinity symbol to indicate going on forever. So we use the notation $\bigcup\limits_{i=1}^{\infty} A_i$ to mean $A_1 \cup A_2 \cup A_3 \cup .....$.

It means what we say it means. It does not include any set labeled $A_\infty$.

And even if there were some set $A_{\infty}$ that is not actually included in the union.

In your case, sure, we can define $I_{\infty} = [0,1]$ if we want (we can define anything we want). But $I_{\infty} \ne I_k$ for any $k$ and $I_{\infty} \not\subset A_1 \cup A_2 \cup A_3 \cup ......... = \bigcup\limits_{n=1}^{\infty}I_n$.

And $\bigcup\limits_{n=1}^{\infty}I_n = (0, 1]$ for the reasons the book gives. $0 \not \in I_k$ for any $k$ so $0 \not \in A_1\cup A_2 \cup A_3.... $

(and for any $x\in (0,1]$ then for any $k > \frac 1x$ we have $\frac 1k < x \le 1$ so $x \in I_k$ so $x \in A_1\cup A_2 \cup ..... \cup A_k \cup A_{k+1} \cup .....$).

Now. We can say $(\bigcup\limits_{n=1}^{\infty} I_n)\cup I_\infty= (I_1\cup I_2 \cup I_3 \cup ........_{\text{it never ends}}....)\cup I_\infty = [0,1]$ but we can't use the notation $\bigcup\limits_{i=1}^\infty A_i$ to mean $(A_1\cup A_2 \cup..... ) \cup A_\infty$ because we already decided that that notation means something else. If we want notation for $(A_1\cup A_2 \cup..... ) \cup A_\infty$ we will have to come up with a different notation.

Addendum: ts;ri (too short; read intensely)

The notation $\bigcup\limits_{i=1}^{n}$ and $\bigcup\limits_{i=1}^\infty$ are actually shorthand. They are shorthand for $\bigcup\limits_{i=1....n}$ and $\bigcup\limits_{i=1....}$ which is to say for $\bigcup\limits_{i\in\{1,2,...., n\}}$ or $\bigcup\limits_{i\in \{1,2,.....\}}$ which is to say $\bigcup\limits_{i\in \mathbb J_n}$ or $\bigcup\limits_{i\in \mathbb N}$.

Which leads to the formal way to do this notation. $\bigcup\limits_{\alpha \in S} A_\alpha$ means that if you have some set $S$ and you index some collection of sets with elements of $S$, then $\bigcup\limits_{\alpha \in S}A_{\alpha}$ would mean the union of these indexed sets.

In this way we have $\bigcup\limits_{k\in \mathbb N} I_k = I_1 \cup I_2 \cup..... = (0, 1]$. While if $I_\infty = [0,1]$ we'd have $\bigcup\limits_{k \in (\mathbb N \cup \{\infty\})} = (I_1\cup I_2 \cup ......) \cup I_\infty = [0,1]$.

That would be the notation you need for what you want to do.

=======

Huge long discussions have come up in discussions so here are some point blank points.

1. "that equation (1) is written with an upper limit of infinity"

That is not what the infinity sign means. The infinity sign is not a limit. The infinity sign is an indication of an indexing of an infinitely long sequence. It only has anything to do when taking actual limits. Otherwise it simply refers to have having a (countably) infinite indexed sequence of things.

So $\sum^{\infty}$ or $\bigcup^\infty$ or $\bigcap^\infty$ means "perform this on all terms". Nothing more, nothing less. It does not mean "perform this all on all terms and on the terms that these approach as a limit to".

You know. Even if you have something like $\lim_{n\to \infty} \frac 1n = 0$ we are not treating the infinity sign as a limit; we are treating the $\lim{}$ symbol as a limit. The $\infty$ sign only means to consider all the items in a sequence. It's the $\lim$ that tells you do take the limit.

So $\bigcup_{k=1}^{\infty} A_i$ means $A_1\cup A_2 \cup A_3 \cup .....$ but you do NOT include $A_{\infty}$. Even if you somehow define $A_\infty = \lim_{n\to \infty} A_n$ (assuming you have somehow defined the "limit" of a sequence of sets), we are not including $A_\infty$ in the union.

2. "Therefore $I_{\infty}=[0,1]$"

Yes, if you define $I_{n} := [\frac 1n, 1]$ and you define $I_{\infty}:=[\lim_{n\to \infty} \frac 1n, 1]$ then, by your definition $I_\infty = [0,1]$.

3. "which contradicts the italicized part"

No. It doesn't. Because $I_\infty$ is not included in the union. Only all the $I_k = [\frac 1k, 1]$ are included.

But the text took it for granted that $I_2\cup I_3 \cup I_4 \cup... = (0,1]$ without explanation. Here is why it follows:

If $x \le 0$ then for any $n\in \mathbb N$ then $x \le 0 < \frac 1n$ and $x \not \in I_n$ for any $n\in \mathbb N$. So $x \not \in \bigcup_{n=1}^\infty I_n$. (Remember: $I_\infty = [0,1]$ is NOT included in the union.

If $0 < x \ge 1$ then $\frac 1x \ge 1$ and there exists and $n\in \mathbb N$ so that $n > \frac 1x$. So $\frac 1n < x \le 1$. So $x\in I_n$. (In fact $x \in I_m$ fore all $m\ge n$). So $x \in \bigcup_{n=1}^\infty I_n$.

If $x > 1$ then for any $n\in \mathbb N$ then $\frac 1n \le 1 < x$ and so $x \not \in \bigcup_{n=1}^\infty I_n$.

So $x \in \bigcup_{n=1}^\infty I_n$ if and only if $0 < x \le 1$ if and only if $x \in (0, 1]$.

So $\bigcup_{n=1}^\infty I_n = (0,1]$ which is not closed.

4. THEOREM. The limit points of $[a,b]$, $(a,b]$, $[a,b)$, and $[a,b]$ are exactly all the the points of $[a,b]$. Therefore $[a,b]$ is closed but $(a,b], [a,b),(a,b)$ are not (they do not all contain both the limit points $a$ and $b$ [but they do contain all the other limit points which are the all the points of $(a,b)$]).

prop: If $w < a$ then $w$ is not a limit point of any of the sets.

Proof. Let $\{w_n\}$ be a sequence that converges to $w$. Let $\epsilon = a-w > 0$. Then there is a $w_k$ so that $|w_k - w| < \epsilon$. This means $w_k < a$. (Because $|w_k - w| < \epsilon \implies -\epsilon < w_k -w < \epsilon \implies w-\epsilon < w_k < w+ \epsilon \implies 2w-a=w-(a-w)=w-\epsilon < w_k < a= w+(a-w) = w+\epsilon$). This means $w_k$ is not in all of the sets.

So there is no sequence that $\{w_n\}$ that converges to $w$ where all the terms are in the sets. So $w$ is not a limit point.

prop: If $w > b$ then $w$ is not a limit point.

Same argument as above. Let $\{w_n\}$ be a sequence that converges to $w$. Let $\epsilon = w-b> 0$. Then there is a $w_k$ so that $|w-w_k| < \epsilon$ and so $-\epsilon< w-w_k < \implies -w-\epsilon < -w_k < -w + \epsilon \implies w-\epsilon < w_k < w+\epsilon \implies b = w-\epsilon < w_k \implies w_k > b$ so there is no that converges to $w$ is in the sets.

prop: If $a < w < b$ then $w$ is a limit point of all sets.

Proof (trivial!). Let $w_k = w$ so the sequence $w_1,w_2,w_3,..... = w,w,w,w....$ trivially converges to $w$. (For any $\epsilon > 0$ and any $n \ge 1$ then $|w_n -w| = |w-w| = 0 < \epsilon$) and $w_k$ is in all the sets. So $w$ is a limit point.

prop: $a$ and $b$ are limit points of all sets.

Let $M = b-a > 0$ and let $u_k = a + \frac M{2^k}$ and $v_k=b-\frac M{2^k}$. Then for all $k\in \mathbb N$ we have $a < u_k =a+\frac M{2^k} < a+M =b$ so $u_k\in (a,b)$ and $u_k$ is in all the sets. And we have $a< w = b-M < b -\frac M{2^k} = u_k < b$. So $v_k$ is in all the sets.

And for any $\epsilon >0$ then for $N \ge \log_2 \frac M{\epsilon}$ we have $n > N \implies 2^n > \frac M{\epsilon} \implies \epsilon > \frac M{2^n}\implies a + \epsilon > a+\frac M{2^n} = u_n > a$. so $|u_n -a| < \epsilon$. So $u_k$ converges to $a$ and $u_k \in (a,b)\subset$ all the sets. So $a$ is a limit point of all the sets.

By the same argument $\epsilon > \frac M{2^n}\implies b-\epsilon < b-\frac M{2^n} = v_n < b$ so $|v_n-a| < \epsilon$. So $v_k$ converges to $b$ and so $b$ is also a limit point of all sets.

Note: even if $w=a$ then $v_k$ is in all sets and $v_k$ converges to $a$ so $a$ is a limit point of all sets (but we can't argue $u_k$ is in $(a,b)$). And even if $w=b$ then $u_k$ is in all sets and $u_k$ converges to $b$ so $b$ is a limit point of all sets (but we can't argue $v_k$ is in $(a,b)$).

We could have combine the arguments for $a\le w \le b$ means $w$ is a limit point of all sets but defining $M = w-a; K = b-w$ and $u_k$ and $v_k$ defined as above and noting that both sequences converge to $w$ and one or the other or both of the sequences are in each of the sets in question. So $w$ is a limit point of all sets even if $w$ isn't in all sets.

Note:

Trivial corollary. If $x \in S$ then $x$ is a limit point of $S$. (By your definition of limit points. Some texts specify a different definition of limit points.)