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We denote by $\mathbb {iN}$ the set of all infinite subsets of $\mathbb N$ and by $f(\mathbb S)$ the set of all finite consecutive subsets of $\mathbb S$. Ex, if $\mathbb S = \{2,4,6,8,\dots \} $ then $f(\mathbb S) = \{ \{2\}, \{2,4\},\{2,4,6\},\{2,4,6,8\},\dots \} $

Then $\mathbb {iN}$ is NOT countable and $\mathbb N,\mathbb S,f(\mathbb S) $are countable

$$Define: \mathbb T = \{ f(\mathbb S) : \mathbb {S} \in \mathbb {iN} \} $$

Here $\mathbb T$ is NOT countable, it has $\mathbb {iN}$ elements

$$Define: \mathbb {U} = \bigcup_{\mathbb {S}\in \mathbb{T}} \mathbb {S} $$

$ \mathbb {U}$ is countable.

$ \mathbb {U}$ is the union of an uncountable number of countable sets.

This would mean that some elements of set $\mathbb T$ contribute information to the union, but most don't.

Why would this be the case if the elements of $\mathbb T$ are homogeneous?

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    $\mathbb{R}$ is uncountable and it is the uncountable union of singletons (finite sets). – 2017-02-17
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    What does $f(\mathbb{S})_\mathbb{S}$ mean? – 2017-02-17
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    The union of a *countably infinite* number of countable sets is countable; but the union of an *uncountably infinite* number of countable sets may be countable or uncountable (for example, if they are all the same set, the union is obviously countable; but if they are all mutually disjoint, the union is uncountable). Just saying "infinite" is too vague. – 2017-02-17
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    Neither $\mathbb{T}$ nor $\mathbb{U}$ are countable.So there is no problem. Btw it would help if you stated clearly where you are defining something. E.g. "Define $\mathbb{T} = \dots$, define $\mathbb{U} = \dots$. – 2017-02-17
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    @Simon Marynissen, you are right, so I changed the title from finite sets to finite objects. $\mathbb{R}$ has singletons, but some of those singletons are infinite objects. – 2017-02-18
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    @Ivan Hieno By definition a singleton is a set with 1 element. – 2017-02-18
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    @Ethan Bolker, I added a dummy variable to make it clear I was using a subscript – 2017-02-18
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    If $S=\{2,4,6,8,\dots\}$ then $\{4,6\}$ is a "finite consecuitive subset" of $S,$ isn't it? – 2017-02-18
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    Is $X_S$ the same thing as $f(S)$? If so, why have two different names for the same thing? – 2017-02-19
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    It's better to say "Disjoint Union" instead of "Union" – 2017-02-19

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In order to ensure that an uncountable union of non-empty sets is uncountable, you need to ensure that there is an uncountable number of sets which actually adds new information to each other.

Look at the function $f(r)=\{0\}$. Then $\bigcup_{r\in\Bbb R}f(r)=\{0\}$. An uncountable union, but a very finite set. How does that happen? Well, we have repetitions.

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    I hate to be difficult as I agree with what you are saying, only we do have an uncountable number of sets which actually do add new information to each other. By your own criteria, set $\mathbb {U}$ should be uncountable, but it is not. – 2017-02-18
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    @IvanHieno "we do have an uncountable number of sets which actually do add new information to each other." That's not quite true! Or at least, it's not true *in the sense that is relevant here*. No $X_S$ has an element which is not in *any other* $X_S$; that is, no $X_S$ has an element which is "unique to itself" (or, *provides information which no other set provides*, to stick with the metaphor above). And so the whole union can be much smaller than you'd predict. – 2017-02-18
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    @Noah Schweber, okay, that makes sense, the only element in $X_S$ that is "unique to itself" is the last element. But, since the last element does not exist, none of the sets have an element that is "unique to itself", therefore most elements are duplicates. Is this the idea that you are thinking of? – 2017-02-19
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    @IvanHieno Pretty much exactly! And this "overlapping information" thing shows up in a lot of ways. One particular weird way: although (exercise) we **cannot** have uncountably many sets of integers which are disjoint (= no element in common), we **can** have uncountably many sets of integers which are *almost disjoint* (= finite intersections). Somehow, allowing nonempty-but-finite intersections gives us "lots of freedom" in creating overlapping information - so much so that we can somehow "fit" uncountably many sets into a very small space. – 2017-02-19
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    @Noah Schweber. I'm having some difficulty understanding the first sentence. If, for instance, we define $R_x$ as $\mathbb{R}-\{x\}$ then$$X=\underset{x\in[0,1]}\bigcup R_x$$ is an uncountable union of non-empty sets, none of which has an element that is not in the union of the rest, but $X$ is uncountable. How would it apply in that case? – 2017-02-19
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    @Martin: The condition is not sufficient *and* necessary. Just sufficient. The question is to what sort of cardinality you can "reduce" your union. In your example, the answer is $2$. In my example, $1$. In the case of subsets of the natural numbers, the answer is countable at best (if not finite). – 2017-02-19
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    The wording would seem to imply "necessary", but if "sufficient" were intended, how would that be relevant to OP's question? – 2017-02-19
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    @Martin: The answer explains that the fact the union is over an uncountable set is not sufficient to conclude that the result is countable, which seems to me as the core confusion in the OP's question. – 2017-02-19
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    OK agreed. (Strictly speaking over the range of a function with an infinite domain; the set of sets involved in the union has cardinality one.) – 2017-02-19
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$\ldots$ if $\mathbb{U}$ is the union of an infinite number of countable sets, then $\mathbb{U}$ should be countable as well $\dots$

This is wrong. Always $X=\bigcup_{x\in X}(\{x\})$ and $\{x\}$ is finite while $X$ may or may not be countable.

Correct is:

$\ldots$ if $\mathbb{U}$ is the union of a countably infinite number of countable sets, then $\mathbb{U}$ should be countable as well $\dots$

$\ldots$ every $f(S)$ should have at least one element in it, that does not appear in any other $f(S)\ldots$

This is also wrong. The mapping $f$ is $1-1$, but any element $x$ of $f(S)$ appears in $f(S')$ for any $S'$ in which $S'\cap\{n\leq\max(x)\}=S\cap\{n\leq \max(x)\}$. (For each $x$ there are $|\mathbb{iN}|$ such $S'$ because the infinite subsets of $\{y\in\mathbb{N}:max(x)

The set $\mathbb{U}$ is just the set of finite subsets of $\mathbb{N}$ which is clearly countable.

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Given any infinite set $\mathbb S_1$ where $ f(\mathbb S_1) = \{ a_1, a_2, a_3, \dots , a_n, \dots \} $

then for any $ a_n $ there will exist an infinite family of infinite sets $\mathbb S_{x \in \mathbb N}$ where $a_n \in f(\mathbb S_x)$

In other words, there would be no member $ a_n $ that is unique to one set only.

Therefore the union of an uncountable number of finite sets would simply be the union of the finite sets, which is countable.