Let $\mathfrak{A}$ be an uncountable collection of Borel sets in $\mathbb{R}^d$ such that for any $A,B\in\mathfrak{A}$, either $A\subset B$ or $A\supset B$. Then is it necessarily true that the union $\bigcup_{A\in\mathfrak{A}}A$ is Borel measurable?
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8$\begingroup$ Every set of size $\aleph_1$ can be written in this way, so "no". $\endgroup$– François G. DoraisCommented Mar 28, 2019 at 3:40
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5$\begingroup$ Since the set has size $\aleph_1$, you can write it as a union of a linearly ordered collection of countable sets. All countable sets are automatically Borel. $\endgroup$– James E HansonCommented Mar 28, 2019 at 5:58
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1$\begingroup$ I am still confused... Say, we have a set $[0,1]$. How do we write it as the union of ordered set $\{A_i:i\in I\}$, in which each $A_i$ is countable? Can you give me an example of these $A_i$? $\endgroup$– user137602Commented Mar 28, 2019 at 6:21
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8$\begingroup$ To complete the answer given by @FrançoisG.Dorais in a comment, note that there is (assuming ZFC) a non-Borel set of size $\aleph_1$. Take a set $X$ of reals of size $\aleph_1$: If the continuum hypothesis holds, then $X$ has more than continuum many subsets of size $\aleph_1$ but there are only continuum many Borel sets. If the continuum hypothesis fails, then $X$ itself isn't Borel because all uncountable Borel sets have the cardinality of the continuum. $\endgroup$– Andreas BlassCommented Mar 28, 2019 at 12:29
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5$\begingroup$ @Skeeve Whether or not CH holds, every uncountable Borel set (in a separable metric space) includes a homeomorphic copy of the Cantor set and therefore has the cardinality of the continuum. (See for example Theorem 13.6 in Kechris's book "Classical Descriptive Set Theory" or Corollary 2C.3 of Moschovakis's book "Descriptive Set Theory".) This holds whether or not CH holds, but it's useful in my earlier comment just when CH fails. $\endgroup$– Andreas BlassCommented Mar 28, 2019 at 21:57
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2 Answers
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Construction
Consider some non-Borel set $Y \subset [0,1]$ (e.g. Vitali set).
Enumerate $Y$ using ordinals as $Y=\{x_\alpha\}_{\alpha < \beta}$. Let $m$ denote the smallest ordinal such that $X=\{x_\alpha\}_{\alpha < m}$ is non-Borel. Note that $m\ge \omega_1$, since otherwise $X$ would be at most countable.
Then the family $\mathfrak A = \{A_\gamma\}_{\gamma<m}$, where $A_\gamma = \{x_\alpha\}_{\alpha<\gamma}$, has the properties desired in the OP, but $\bigcup_{A \in \mathfrak A} A$ is non-Borel.
For more details about ordinals see e.g. Set theory by T. Jech (2006).
Discussion of cardinality
Let us prove that in fact $|X| = \aleph_1$.
Indeed, by construction $\aleph_1 \le |X| \le 2^{\aleph_0}$. Any uncountable Borel set has cardinality $2^{\aleph_0}$ (e.g. by Theorem 13.6 in Classical Descriptive Set Theory by A.S. Kechris or Corollary 2C.3 in Descriptive Set Theory by Y.M. Moschovakis.). Therefore, if CH fails then $|X|\le \aleph_1$, since otherwise there would exist a Borel subset with cardinality $\aleph_1$ (by minimality of $m$). On the other hand if CH holds then immediately $|X| = \aleph_1$.
Most of this answer comes from the very useful comments in this thread.
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4$\begingroup$ You seem to use the inequality $2^{\aleph_0}<2^{\aleph_1}$, also known as "weak CH". This is not provable in ZFC. - Also, I don't think you can prove from ZF (without AC) that the reals contain a set of cardinality $\aleph_1$. $\endgroup$ Commented Mar 28, 2019 at 11:10
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4$\begingroup$ Also, without choice, it is consistent that every set of reals is Borel. $\endgroup$ Commented Mar 28, 2019 at 11:43
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2$\begingroup$ Why don't you just take a non-Borel set of minimum cardinality and well-order it so all initial segments have lower cardinality? That just needs AC, no CH at all. $\endgroup$– bofCommented Mar 28, 2019 at 12:38
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1$\begingroup$ The comment of Andreas Blass on the question gives another argument that, in ZFC, there exists a non-Borel set of cardinality $\aleph_1$. Again, no CH needed. $\endgroup$ Commented Mar 28, 2019 at 12:51
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1$\begingroup$ @bof thanks a lot for the suggestion, I've updated my answer using it. $\endgroup$– SkeeveCommented Mar 28, 2019 at 21:52
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Every coanalytic set can be written as a union of $\aleph_1$-many Borel sets. (For example, see Theorem 4.3.17 of Srivastava.) Take a coanalytic non-Borel set $B \subseteq \mathbb{R}^d$. Then $B = \bigcup_{\alpha < \omega_1} B_{\alpha}$ for some Borel sets $B_{\alpha}$. For each $\alpha < \omega_1$, set $C_{\alpha} = \bigcup_{\delta < \alpha} B_{\delta}$. Clearly each $C_{\alpha}$ is Borel since it is a countable union of Borel sets. On the other hand, $\bigcup_{\alpha < \omega_1} C_{\alpha}=\bigcup_{\alpha < \omega_1} B_{\alpha}=B$ is non-Borel.
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$\begingroup$ I see. Thank you for the answer! $\endgroup$ Commented Mar 29, 2019 at 7:05