Linked Questions

2
votes
6answers
650 views

Closed, uncountable set without rationals. [duplicate]

Problem: Find a closed, uncountable without rationals, subset in the unit interval. Solution: Let X be any binary irrational number , replace the 1s with 2s in the decimal expansion . Now we can ...
3
votes
1answer
4k views

Perfect set in $\mathbb{R}$ which contains no rational number [duplicate]

Possible Duplicate: Perfect set without rationals Does there exist a nonempty perfect set in $\mathbb{R}$ which contains no rational number? This problem is on p.44 PMA - Rudin I found a proof ...
3
votes
2answers
742 views

Perfect set without rational numbers [duplicate]

Sorry if this problem is repeated. Is there a nonempty perfect set in $\mathbb{R}^1$ which contains no rational number? Proof sketch: This set must be uncountable because any nonempty perfect set in ...
1
vote
1answer
559 views

Rudin exercise 2.18: perfect set with only irrational numbers. [duplicate]

I am trying to solve exercise number 2.18 in Rudin. The question is the following: "Is there a nonempty perfect set in R1 that contains no rational numbers"? My solution is basically to construct a ...
3
votes
1answer
631 views

Rudin's PMA Exercise 2.18 - Perfect Sets [duplicate]

I've been working through Chapter 2 questions and have thought about Exercise 2.18 for a while, but couldn't come up with an answer. Is there a nonempty perfect set in R which contains no rational ...
1
vote
0answers
46 views

Is there a non-empty subset of $\mathbb{R} $ like $A$ such that the set of accumulation points of $A$ is itself and $A\cap\mathbb{Q} = \emptyset $ [duplicate]

Is there a non-empty subset of $\mathbb{R} $ like $A$ such that the set of accumulation points of $A$ is $A$ and $A\cap\mathbb{Q}=\emptyset\,$?
0
votes
0answers
40 views

Non empty Perfect set with only irrational numbers [duplicate]

I encountered this question in Abbott's Understanding Analysis. The problem asks to construct a nonempty perfect set with no rationals. It starts with enumerating the rationals $\mathbb{Q}=\{r_1,r_2,.....
0
votes
0answers
16 views

Basic Topology about metric space $R$ [duplicate]

Is there a nonempty perfect set in $R$ which contains no rational number? Its the excercise of Principles of Mathematical Analysis by Walter Rudin. I can't find the answer.I try to construct a set as ...
13
votes
3answers
2k views

Irrational Cantor set?

Can someone help me? How can I prove that exists a number $k \in \mathbb R$ that $$A = \{x + k;\ x \in \text{Cantor set} \} \subset\text{ Irrationals}\;?$$
10
votes
4answers
708 views

Questions about open sets in ${\mathbb R}$

Consider the following problem: Let ${\mathbb Q} \subset A\subset {\mathbb R}$, which of the following must be true? A. If $A$ is open, then $A={\mathbb R}$ B. If $A$ is closed, then $A={\mathbb R}$...
8
votes
1answer
2k views

Is there a compact subset of the irrationals with positive Lebesgue measure?

Does there exist $K \subseteq \mathbb{R} \backslash \mathbb{Q}$ such that $K$ is compact, and has Lebesgue measure greater than $0$? As I have been trying to think of examples, I suspect that any ...
8
votes
2answers
283 views

Is it possible that all subseries converge to irrationals?

Does there exists a positive decreasing sequence $\{a_i\}$ with $\sum_{i\in\mathbb{N}} a_i$ convergent, such that $\forall I\subset\mathbb{N},\sum_{i\in I}a_i$ is an irrational number? Such an ...
4
votes
4answers
856 views

Discontinuous function at an uncountable set with not rationals

Does there exists a function $f:[0,1]\to\mathbb{R}$ such that $D(f)$ (its points of discontinuity) is an uncountable set containing no rational number? First thing I thought of was $\mathbb{R}\...
3
votes
3answers
259 views

Question about irrational numbers

How to show : If $A \subseteq \mathbb{I}$ is not countable, then $A$ is not closed set in $\mathbb{R}$. I think that if $A$ is closed, then $A^{c}$ is open thus: $A^{c}$ is an union of open ...
3
votes
1answer
256 views

Why is this set of continued fractions perfect?

Would somebody please explain why the set of continued fractions in this answer https://math.stackexchange.com/a/1067/20873 i.e. "the set of all irrationals with continued fractions consisting only ...

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