Intereting Posts

Continuous bijection from $(0,1)$ to $$
Basis for the Riemann-Roch space $L(kP)$ on a curve
$\Delta u$ is bounded. Can we say $u\in C^1$?
How to pass from $L^2(0,T;V')$ to $\mathcal{D}'\big(\Omega\times (0,T)\big)$?
Show that the set of functions under composition is isomorphic to $S_3$
Unique expression of a polynomial under quotient mapping?
Singular value decomposition proof
Is this reasoning correct? Connection with torsion on SO(3)
Identity in Number Theory Paper
How to derive the gregory series for inverse tangent function?
Proving Cantor's theorem
How should I calculate the $n$th derivative of $f(x)=x^x$?
Integration of $\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$
Affinely Extended Reals as a metric space
Concecutive last zeroes in expansion of $100!$

Since the construction of a partition of two Bernstein sets is almost identical to that of a partition of three in an uncountable Polish space. It’s possible that the union of a Bernstein set and a singleton is a Bernstein set. But is it possible otherwise, that is, the union of a Bernstein set and a singleton contains an uncountable closed set ?

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- Quotient space of the reals by the rationals
- Question about limit points of a Subset of $\mathbb{R}$
- When is the image of a proper map closed?
- Set of limit points of a subset of a Hausdorff space is closed.
- an example of an interesting connected topological space
- Map from binary sequences on $\{0,1\}$ into the Cantor set $C$ respects order

No.

To see this, notice that if you remove a point from a perfect set, what remains will still contain a perfect set, so the intersection of a Bernstein set with a perfect set is not just nonempty, but also infinite. In fact, it’s dense (in the perfect set) by a similar argument.

Therefore, removing (or adding) a single point does not turn a Bernstein set to a non-Bernstein set.

Using measure arguments you can show that the intersection is uncountable: the intersection of a Bernstein set with any perfect set has full outer measure with respect to any continuous probability measure on the perfect set, so it can’t be countable.

(I vaguely recall that any Bernstein set is a union of $\mathfrak c$-many Bernstein sets, so that the intersection is actually of cardinality of the continuum, but I don’t remember the proof, exactly, and might be mixing something up.)

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