Changing the lengths of the intervals excluded during the construction of the ternary Cantor set, show that is possible to build a compact, totally disconnected and perfect set (a Cantor set) with positive Lebesgue measure.   I have no idea on how to solve this question. It seems also counter-intuitive to me there is a Cantor set whose Lebesgue measure is positive.   Question:  How should I solve the exercise?   Thanks in advance!
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So the standard Cantor set has an outer measure equal to $0$, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of $1$. I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of $0$ as well...  Are there other constraints that need to be made in order to accomplish this?
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I found in a book a proof that the Cantor Set $\Delta$ is perfect, however I would like to know if "my proof" does the job in the same way.     Theorem: The Cantor Set $\Delta$ is perfect.        Proof:   Let $x \in \Delta$ and fix $\epsilon &gt; 0$. Then, we can take a $n_0 = n$ sufficiently large to have $\epsilon &gt; 1/3^{n_0}$.   Thus, the interval $[a, b]$ where $x$ lies is a subset of $B_\epsilon &gt; (x)$. Hence, by iterating the construction of the Cantor set for $N &gt;  n_0$, we have intervals of length $1/3^N$ all included in $B_\epsilon  (x)$, but with only one of those intervals such that $x$ lies within.   The intution behind the proof was that we should prove that for every $x$, if $x \in \Delta$, then for every $\epsilon &gt;0$, $B_\epsilon (x) \setminus \{x\} \cap \Delta \neq \varnothing$.    Now, I do not particularly like my reference to the $[a, b]$ interval that is not mentioned before. Moreover, here – by choosing a closed interval – I am trying to address all at once the case in which $x$ is an endpoint of one of the closed intervals that form $\Delta$. Finally, I did not close the proof with a statement like "Thus, there are infinitely many points that differ from $x$ and that lie within $B_\epsilon (x)$.    In the end, I am not completely sure if this can be considered a proof or not. The intuition is correct (I am kind of positive about it), but I am not sure if I was actually able to write down my intuition in a good way.    As always any feedback is more than welcome. Thank you!
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As can be seen  (exercise 1), Terry Tao gives an exercise to find the Minkowski Dimension of the Quadnary Cantor Set, and of a special Quadnary Cantor Set.  The two sets are: $$C := \left\{\sum_{i=1}^{\infty} a_i4^{-i} : a_i \in \{0,3\} \right\}$$  and   $$C' : = \left\{\sum_{i=1}^{\infty} a_i4^{-i} : a_i \in \{0,3\} \, \, \text{if} \, \, (2k)! \leq i \leq (2k+1)! \,\, \text{and arbitrary otherwise} \right\}.$$  I managed to find the Minkowksi Dimension of the first one by noting that $C_\delta$, the $\delta$-fattening of $C$, has $\text{vol}(C_\delta)$ equal to the $n^{th}$ iteration of the Quadnary Cantor set.    For the second one, I've found a closed form expression for the volume if $\delta_n = 4^{-(2n)!}$, since I suspect this should be a lim sup, but I'm not sure.   The closed form I found is:  $$\text{vol}(C_{\delta_n}) = \sum_{k=1}^{n-1} \sum_{j=0}^{2k(2k)!} 2^{-(2k-1)!+2j)},$$  but I'm not sure how to even begin manipulating the expression  $$ \frac{\log(\text{vol}(C_{\delta_n}))}{-(2n)! \log(4)}. $$  In particular, we are to show that the upper MD of $C'$ is $1$, while the lower MD is $\frac{1}{2}$, but it seems to me that the worst sequence we can use is $4^{-(2n+1)!}$, since this would give a volume of  $$\text{vol}(C_{\delta_n}) = \sum_{k=1}^{n} \sum_{j=0}^{2k(2k)!} 2^{-(2k-1)!+2j)}.$$  To note how I got my closed form expression, at the $i=(2k)!$ iteration, we are removing  $$2^{(2k)! + ((2k)! - (2(k-1)+1)!)}$$ intervals of length $$4^{-(2k)!}$$ note that we must remove $((2k)! - (2(k-1)+1)!)$ more intervals, since we did not remove anything for some time and also that $$(2k+1)! - (2k)! = 2k(2k)!.$$  Also it is clear that the lim inf (lower MD) is at least $\frac{1}{2}$, since $C \subset C'$, but it's unclear why it isn't strictly larger.
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I was thinking if there is a solution to this problem without trying to explicitly create Vitali sets in $A$. Does anyone have any ideas?
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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 $\mathbb{R}^k$ is uncountable.   We know that Cantor set (C) is perfect set on a line. If we take $E=C+\sqrt{2}=\{x+\sqrt{2}: x\in C\}$. Will this set be an answer to above problem?
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Let $\gamma : [0,1]\rightarrow\mathbb R^2$ be a continuous curve whose image has positive Lebesgue measure. Must $\gamma$ have self-intersections? Intuitively this seems like it should be true, but I could not find a proof. An equivalent question is whether there is a positive measure subset of $\mathbb R^2$ which is homeomorphic to $[0,1]$.   Note that it is shown here  that $\gamma$ must in fact have lots of intersections if the image of $\gamma$ is a rectangle. But, here I am not making any topological assumptions on the image of $\gamma$. The answer to the linked question does not work in this generality.
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Ever since day one of of my Mathematical Logic course, this fact has really bothered me. I cannot wrap my head around how an empty set is a subset of every possible set. Could someone kindly explain how this is true? Any help is appreciated!
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I'm doing exercise from "Topology without tears book", page 27, exercises a and b.   First one (a) reads as follows. "Let $\mathbb{R}$ be the set of all real numbers. Prove that each of the following collections of subsets of $\mathbb{R}$ is a topology.  a. $\tau_1$ consists of $\mathbb{R}$ , $\emptyset$ and every interval $(-n;n)$ for $n$ any positive integer  b. $\tau_1$ consists of $\mathbb{R}$ , $\emptyset$ and every interval $[-n;n]$ for $n$ any positive integer"  To prove this I noticied that     $S_n=(-n;n) \in \tau_1 \; \forall n \in \mathbb{N}$ by definition.   $S_{n_1} \subset S_{n_2} \; \forall \; n_1 &lt; n_2, n_1,\;n_2 \in \mathbb{N}$   From 2 it follows that    $S_{n_1} \cap S_{n_2} = S_{n_1} \; \forall n_1 &lt; n_2 $  $\bigcup_{i=1}^{i=n} S_i = S_n \in S_{n+1} $   Sience intersection of any two sets is in $\tau_1$ and union of any number of sets is also in $\tau_1$ and noticing that $S_n \subset \mathbb{R} \; \forall n \in \mathbb{N}$ I can state that $\tau_1$ is a topology of $\mathbb{R}$  Is this proven right? I have the following question. Exercise b is very similar to exercise a - the only difference is that now $[-n; n]$ is a closed interval. I think that I still can use the same prove for this case too, but not sure - does closed interval $[-n;n]$ changes anything?
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First, this question is motivated by the imprecise question: Is there a sensible notion of parity (evenness and oddness) for real numbers?   Here are some properties a notion of parity for $\mathbb{R}$ should have:    It should be an equivalence relation on $\mathbb{R}$ with exactly two equivalence classes.  Each equivalence class should be dense in $\mathbb{R}$. There should be some kind of symmetry between the two equivalence classes. (This is intentionally imprecise.)   If we divide $\mathbb{R}$ into rationals and irrationals, this lacks symmetry, since the rationals have measure zero and are countable, while the rationals have infinite measure and are uncountable.   So we could require that each equivalence class be uncountable and/or have positive (or infinite) measure. Then take one equivalence class to be the rationals, union with countably many fat cantor sets with positive measure, and the other class its complement. Then both classes are now dense, uncountable, and have infinite measure. But again, symmetry is lacking. One of these classes is "uncountable everywhere" and has "positive measure everywhere" while the other does not.   I propose some definitions: A subset $A \subset \mathbb{R}$ is uncountable everywhere if for any open interval $(a,b)$, the intersection $A \cap (a,b)$ is uncountable. A subset $A \subset \mathbb{R}$ has positive measure everywhere if for every open interval $(a,b)$, the intersection $A \cap (a,b)$ has positive measure. We can see immediately that having positive measure everywhere implies being uncountable everywhere, since a positive measure set must be uncountable.   Finally, my questions:    Is there a partition of $\mathbb{R}$ into two sets that are both uncountable everywhere?  Is there a partition of $\mathbb{R}$ into two sets that both have positive measure everywhere?  Is there a partition of $\mathbb{R}$ into two sets that split each interval $(a,b)$ into two parts of equal measure?   Potential generalizations for extra credit: What about partitions of $\mathbb{R}$ with $n$ equivalence classes, where $n \in \mathbb{N}$, or even with countably many, or even uncountably many equivalence classes?  Note: If you've seen these definitions of uncountable everywhere or positive measure everywhere somewhere under a different name, please let me know. I've never found anything where other people were thinking about these notions.  EDIT: By "measure," I mean Lebesgue outer measure, so we don't have to worry about anything being measurable.
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Let $\mathcal A$ be a collection of pairwise disjoint subsets of a $\sigma$-finite measure space, and suppose each set in $\mathcal A$ has strictly positive measure. Show that $\mathcal A$ is at most countable.  I have no clue about this one. Any hints would be appreciated.
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Theorem. Let $(X,\mathcal B,\mu)$ a finite measure space, where $\mu$ is a positive measure. Let $\mathcal A\subset \mathcal B$ an algebra generating $\cal B$. Then for all $B\in\cal B$ and $\varepsilon&gt;0$, we can find $A\in\cal A$ such that $$\mu(A\Delta B)=\mu(A\cup B)-\mu(A\cap B)&lt;\varepsilon.$$  I don't think there is a proof in this site. It's a useful result for several reasons:  We know what the algebra generated by a collection of sets is, but not what the generated $\sigma$-algebra is. The map $\rho\colon \cal B\times\cal B\to \Bbb R_+$, $\rho(A,A')=\mu(A\Delta A')$ gives a pseudo-metric on $\cal B$. This makes a link between generating for an algebra and dense for the pseudo-metric. We say that a $\sigma$-algebra is separable if it's generated by a countable class of sets. In this case, the algebra generated by this class is countable. An with the mentioned result, we can show that $L^p(\mu)$ is separable for $1\leq p&lt;\infty$, which makes a link between the two notions. In ergodic theory, we have to test mixing conditions only an a generating algebra, not on all the $\sigma$-algebra.
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Give an example to show that Cantor's Intersection Theorem would not be true if compact sets were replaced by closed sets.    Compact set is closed and bounded, so what I'm going to find is something that is closed but not bounded.  By the Cantor theorem, which says that a decreasing sequence of non-empty compact subset will have a non-empty intersection.  What I was trying to approach is how two sets just "touched" and as if they have not bounded, two sets "touched" nothing. But I cant give an example for that
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The Cantor set, as showed in books and , is defined in terms of $C_k$, the finite Cantor set of level $k$:  $$ \mathcal{C} = \bigcap_{k=1}^\infty C_k $$  But after intersection only the "last" remains, so, why not to define it by a limit?    $$\mathcal{C} = \lim_{k \to \infty} C_k$$  It is perhaps a naive intuition, but I not see a good justification.    (adding here a note after first answer, only for comment the comments)  NOTE: if it is not only a  question of choice of notation, but also about context and semantics. Can I tell an engineer that the intersection is a kind of specification, something like a project to explain "what I need", and the limit is a "what  I get", the end result?&nbsp; &nbsp; ...Or perhaps the inverse, as suggested by @HansLundmark (thanks the comment! also thnaks @SangchulLee!). I am supposing that $C_k$ is a "decreasing sequence",  $C_1 \supseteq C_2 \supseteq C_3 \supseteq \dotsb$, so using  we can say  &nbsp;"it's natural to define the limit as the intersection: $C_n \to \mathcal{C}$ as $n\to\infty$",  &nbsp;where $\mathcal{C}$ is defined by intersection.  About comment of @LordShark: its is possible to use  "limits" notation in the context of set sequences without "develop a theory" for it?  @HansLundmark's link is a satisfactory answer for it?
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Given a positive real number construct a non measurable set of that outer measure   I've been asked this question and am not able to construct it.  I am well aware of the construction of a Vitali set, and I know that their outer measure are not known explicitly.  I can construct a set dense in any interval by taking the union of the standard non measurable set with rationals of that interval but that doesn't serve my purpose because measure of countable set is zero and doesn't do anything good to the required result.
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This is an interesting question that has stumped the entirety of my measure theory class, including the professor:  Prove or disprove:  Let $(X,\mathcal A)$, $(Y,\mathcal B)$ be measure spaces.  Let $f$ be an $\mathcal A$ $\times$ $\mathcal B$ measurable nonnegative function, and let $g(x)$ $=$ $sup_{y \in Y}f(x,y)$, with $g(x)&lt;\infty$ for all $x$. Is $g(x)$ necessarily an $\mathcal A$-measurable function?  We all feel the answer is no, given that slices are measurable, and sups of measurable functions are only guaranteed to be measurable over a countable index. We think the correct answer is to start with a nonmeasurable set $S$ in $X$, and to try to build a set $T$ in $Y$ that makes $S \times T$ measurable in $\mathcal A \times \mathcal B$, but we suspect this is quite difficult with no further guidance.   Any ideas?
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$m$ is the Lebesgue measure. I was thinking that: $\Bbb Q$ is dense and $m(\Bbb Q)=0&lt;1$ but it fails to be open, but maybe I could construct an open set from this fact, or also using that $(0,1)$ is a dense and open subset but $m(0,1)=1$.   Any hints or ideas will be very appreciated. Thanks.
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Is it possible to construct a non-measurable set in $[0,1]$ of a given outer measure $x \in [0,1]$? This will probably require the axiom of choice. Does anyone have a suggestion?  Edit: I forgot to mention that the set has to be in $[0,1]$ which seemed "clear" in my head, but which obviously isn't.  Edit 2: Now it is clear that I just need a non-measurable set with outer measure 1, from that I can make any other one. I have once seen somewhere that they do this with transfinite recursion (if I recall correctly), but I don't remember where and how because I didn't know that transfinite recursion was back then.