3
votes
2answers
75 views

Showing a set is nowhere dense in $C([0,1])$

Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that \begin{align*} \lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert, ...
0
votes
1answer
27 views

A set that is a countable intersection of open and dense sets but not open.

We know that Baire Category Theorem implies that in a complete metric space, the countable intersection of open and dense sets is nonempty and actually dense itself. But it is clear that a countable ...
3
votes
1answer
115 views

Prove it doesn't exist any function f:R→R that is continuous only at the rational points.

Prove it doesn't exist any function $f:\mathbb R \to \mathbb R$ that is continuous only at the rational points. Suggestion: For every $n \in \mathbb N$, consider the set $U_n=\{x \in \mathbb R : ...
1
vote
1answer
216 views

Proving the existence of a non-monotone continuous function defined on $[0,1]$

Let $(I_n)_{n \in \mathbb N}$ be the sequence of intervals of $[0,1]$ with rational endpoints, and for every $n \in \mathbb N~$ let $E_n=\{f \in C[0,1] : f \:\text{is monotone in}\: I_n\}$. Prove that ...
4
votes
1answer
137 views

$L_2$ is of first category in $L_1$ (Rudin Excercise 2.4b) [duplicate]

We mean here $L_2$, and $L_1$ the usual Lebesgue spaces on the unit-interval. It is excercise 2.4 from Rudin. There's several ways to show that $L_2$ is nowhere dense in $L_1$. But in (b) they ask to ...
4
votes
1answer
153 views

Prove $\ell_1$ is first category in $\ell_2$

Prove that $\ell_1$ is first category in $\ell_2$. I tried to solve this, but had no idea about the approach. Any suggestions are helpful. Thanks in advance.
2
votes
2answers
548 views

Corollary to Baire's Category Theorem

In Rudin's Real and Complex Analysis (p. 97 in my 3rd edition), the following is stated as a corollary to Baire's Category Theorem: "In a complete metric space, the intersection of any countable ...
44
votes
5answers
5k views

Is $[0,1]$ a countable disjoint union of closed sets?

Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?