digital signature – Proving authenticity of a message from a message app in case of deletion

Say you want to prove that you received a certain message from someone.
This can be difficult because many messaging apps (like facebook messenger) allow the sender to delete messages on the recipient’s end.

Of course, you could screenshot the message, but then you still need to prove 2 things:

  1. The time of the message.
  2. The authenticity of your screenshot.

1 is easy, if you upload a hash to public databases
as described in this question: Proving creation time/date of a screenshot

2 is a bit more difficult. You’d have to prove somehow that you didn’t just forge the screenshot. Note: this is a little different than proving that it was the other person that sent it. This is proving that you received the message from a certain account.

Is there a way establish proof of this?

Proving an algorithm satisfies $epsilon$-DP

I am trying to prove/disprove that an algorithm satisfies $epsilon$-DP. I proved some of them but there are 3 more which I could not decide on.

Here they are:

  • Algorithm A takes as input a dataset D and counts the number of records in D. If the number of records is greater than eε, A prints out: “large”. Otherwise, A prints out “small”.
  • Algorithm A takes as input a dataset D and an integer k. It returns TRUE if D is k- anonymous and FALSE otherwise.
  • Algorithm A wants to answer if a certain individual’s age is above 40. To do so, A takes as input dataset D and a person’s name (eg: “John Doe”) as input. It retrieves from D the count of individuals with that name and age > 40. Then, A adds Laplace noise with mean = 0 and scale = ε. If the noisy count is > 1, A returns TRUE. Otherwise, A returns FALSE.

If anybody can help me to solve them, I would appreciate a lot.

analytic number theory – Do we have any result proving a strong upper bound on the cardinality of set $P_{alpha}(x)$ for some large parameter $x$?

Define $x_0=0$ and $x_{i+1} = P(x_i)$ for all integers $i ge 0$.
Let $l(p)$ to be the least positive integer such that $p|x_{l(p)}$ for some prime $p$.

Then if we let
$$P_{(0<alpha<1)} = { pin mathbb{P}mid l(p)<p^{alpha}}$$ where $mathbb{P}$ is the set of primes, do we have any result proving a strong upper bound on the cardinality of set $P_{alpha}(x)$ for some large parameter $x$?

wallet – Proving two public keys are derived from the same seed, without revealing the xpub and the path?

Let’s think about the following scenario:

  1. Key0 is derived from SEED at path m/44’/0’/0’/0/0
  2. Key1 is derived from SEED at path m/44’/0’/0’/0/1

I am wondering if there’s a way to prove that Key0 and Key1 share the same seed WITHOUT having to share the SEED or any Xpub along the way.

Basically I want to be able to prove two keys are somehow related without risking security (Where if you share an Xpub and one of its child private keys gets leaked, the entire sibling set is compromised).

It doesn’t have to be exactly this, but my goal is to prove that two keys are related without sacrificing security, so if there are ways to do this, I would appreciate those solutions too.

algorithm analysis – Proving that the recurrence $T(n) = 2Tleft(frac{n}{2}right) + 1$ with $T(2) = 1$ is asymptotically $O(n)$

I’ve already solved the recurrence exactly and found that $T(n) = n – 1$. Therefore, I know that $T(n) = O(n)$.

However, I’m having trouble showing that $T(n) = O(n)$ without solving the recurrence exactly.

My strategy so far has been to prove that $T(n) leq cn$ ($c$ is a constant) $forall n geq 2$ via induction on $n$.

Clearly, the base case holds $T(2) = 1 leq 2c$ as long as $c geq frac{1}{2}$.

For the inductive hypothesis (IH), I assume $forall k < n$, $T(k) leq ck$.

Finally, for the inductive step, I have

$$
begin{alignat*}{2}
T(n) &= 2Tleft(frac{n}{2}right) + 1 &&text{ (by definition)}\
&leq 2cleft(frac{n}{2}right) + 1 &&text{ (by IH)}\
&= cn + 1
end{alignat*}
$$

However, I can’t conclude that $T(n) leq cn$ from this. Where am I going wrong?

probability – Proving inequalities using Markov’s

Let $$X_1,…,X_n$$ be independent Bernoulli random variables
with parameters $$p_1,…,p_n$$
$X = X_1 +….+X_n$

How do I show that
$$Pr(X geq sqrt{2} + sum_{i=1}^{n}p_i) leq sum_{i=1}^{n} frac{p_i(1-p_i)}{2}$$

I know that $$sum_{i=1}^{n}p_i = sum_{i=1}^{n}E(X_i) = E(X)$$
So we can replace $Pr(X geq sqrt{2} + E(X))$ and using markov we get $$frac{E(X)}{sqrt{2} + E(X)}$$

I would like to know how to show the above is less or equal to what we want to prove?
$$frac{E(X)}{sqrt{2} + E(X)} leq sum_{i=1}^{n} frac{p_i(1-p_i)}{2}$$

real analysis – Proving weak convergence for a specific function.

My argument (briefly):

Using density of the simple functions in $textit{L}^{infty}((0,1))$, i must prove the definition for weak converges holds, but i can restrict the argument to simple functions.

So given a simple function $overline{g}$ i know there exist costants $c_1=0<c_2<cdots<c_n=1$ such that $overline{g}$ is costant on these, so you can come to this partial conclusion:
$$lim_{nto +infty} int_{(0,1)}overline{g}u_ndm=lim_{nto +infty}sum_{j=1}^{k-1}overline{g}(c_i)int_{c_j}^{c_{j+1}}u_n(z)dz=sum_{j=1}^{k-1}overline{g}(c_j)lim_{nto +infty}int_{c_j}^{c_{j+1}}u_n(z)dz=$$
$$= sum_{j=1}^{k-1}overline{g}(c_j)biggl( lim_{nto +infty}int_{(0,1)}achi_{E_n^{(j)}}(z)dz + lim_{nto +infty}int _{(0,1)}bchi_{F_n^{(j)}}(z)dz biggl) =$$
$$= (star)sum_{j=1}^{k-1}overline{g}(c_j)biggl( a lim_{nto +infty}|E_n^{(j)}| + b lim_{nto +infty}|F_n^{(j)}| biggl)(star)$$
where
$$E_n^{(j)}:=biggl{{} bigcup_{ile n-1}biggl(frac{i}{n},frac{i+1/2}{n} biggr) bigcap (c_j,c_{j+1}) biggr{}} mbox{ e } F_n^{(j)}:=biggl{{} bigcup_{ile n-1}biggl(frac{i+1/2}{n},frac{i+1}{n} biggr) bigcap (c_j,c_{j+1}) biggr{}} mbox.$$

These next step is where i think i put to much effort for a proof like this, but i am simply finding the limit of the measure of these two sets.

Lets take $I_n^{(j)}:={ ile n-1:bigl(frac{i}{n},frac{i+1}{n}bigr)subseteq (c_j,c_{j+1}) }$, by definition
$$(star star)E_n^{(j)} supseteq bigcup_{iin I_n^{(j)}}biggl(frac{i}{n},frac{i+1/2}{n}biggr) mbox{ e } F_n^{(j)} supseteq bigcup_{iin I_n^{(j)}}biggl(frac{i+1/2}{n},frac{i+1}{n}biggr) (star star)mbox,$$

and we can observe that for $n ge frac{2}{(c_{j+1}-c_j)}$ we get
$$bigcup_{l=0}^{lfloor n(c_{j+1}-c_j)rfloor-1} biggl( c_j+frac{l}{n},c_j+frac{l+1}{n} biggr) mbox{, with measure } le |(c_j,c_{j+1})|$$
and $exists$ $overline{i}in I_n^{(j)}$ such that
$$frac{overline{i}}{n}in biggl(c_j,c_j + frac{1}{n}biggr)mbox,$$
We can observe that we can obtain our sets
$$ bigcup_{iin I_n^{(j)}}biggl(frac{i}{n},frac{i+1}{n}biggr) mbox,$$
by a translation $le 1/n$ to the right of the sets $bigcup_{l=0}^{lfloor n(c_{j+1}-c_j)rfloor-1} biggl( c_j+frac{l}{n},c_j+frac{l+1}{n} biggr)$. With this process i might loose the last of my intervals, but I loose at most one.

So i can deduce this: $|I_n^{(j)}| ge lfloor n(c_{j+1} – c_j) rfloor -1 ge n(c_{j+1}-c_j)-2$.

Now, using $(star star)$,

$$lim_{nto +infty}|E_n^{(j)}| ge lim_{nto +infty}sum_{iin I_n^{(j)}}biggl| biggl( frac{i}{n},frac{i+1/2}{n} biggr) biggr| = lim_{nto +infty}sum_{iin I_n^{(j)}}frac{1}{2n} = lim_{nto +infty}frac{|I_n^{(j)}|}{2n} ge$$
$$ge lim_{nto +infty}frac{n(c_{j+1}-c_j)-2}{2n} = frac{(c_{j+1}-c_j)}{2}$$
and in the same way
$$lim_{nto +infty}|F_n^{(j)}| ge frac{(c_{j+1}-c_j)}{2} mbox,$$
but since we have also that $|E_n^{(j)} cup F_n^{(n)}|=|E_n^{(j)}|+|F_n^{(j)}|=|(c_j,c_{j+1})|=(c_{j+1}-c_j)$, we can deduce that
$$lim_{nto +infty}|E_n^{(j)}|=lim_{nto +infty}|F_n^{(j)}|=frac{(c_{j+1}-c_j)}{2} mbox.$$

This holds $forall 1ge jge k-1$, so now if we take
$$nge max_{j}biggl{{} frac{2}{(c_{j+1}-c_j)} biggr{}}$$
we can conclude our first series of equalities in $(star)$ with:
$$=lim_{nto +infty}int_{(0,1)}overline{g}u_ndm = sum_{j=1}^{k-1} overline{g}(c_j)frac{(a+b)}{2}(c_{j+1}-c_j)=sum_{j=1}^{k-1}int_{c_j}^{c_{j+1}}overline{g}(z)frac{(a+b)}{2}dz=int_{(0,1)}overline{g}(z)frac{(a+b)}{2}dz mbox,$$
which is exactly what we wanted.

proof writing – An attempt at proving that $A=(0,1)$ is not compact on the real line with the usual topology.

I am supposed to show that the open interval $(0,1)$ on the real line $mathbb{R}$ (with the usual topology) is not compact. I know that one of the examples of an open cover without a finite subcover, in this case, would be $A=left{left(dfrac{1}{n},1right)biggrvert ninmathbb{N}right}$, thus proving that $A$ is not compact.

However, I am trying to prove the same using a different example of an open subcover, $$mathscr{F}=left{left(dfrac{1}{2^n},dfrac{3}{2^n}right)biggrvert ninmathbb{N}right}.$$ If it can be shown that the family of sets $mathscr{F}$ forms an open cover for $(0,1)$, i.e., $bigcuplimits_{ninmathbb{N}}G_n$, where $G_n=left(dfrac{1}{2^n},dfrac{3}{2^n}right)$, it is straightforward to check that for any $x=dfrac{1}{2^n}$, where $ninmathbb{N}$, it only lies in one of these $G_n$‘s. So, removing any one of these open sets from our open cover, in an attempt to find a finite subcover, the family of sets won’t form a finite subcover of $(0,1)$. Thus, $(0,1)$ is not compact.

Where I am getting stuck is in actually formally proving that any $xin(0,1)$ lies in one of these $G_n$‘s. My intuition here is that by the use of Archimedean principle, for any $xin(0,1)$, there exists some $kinmathbb{N}$ such that $x>dfrac{1}{k}$. So, let us choose $k$, for a given $x$, such that $dfrac{1}{k}<x<dfrac{1}{k-1}$. Again, since this $k>1$, there exists some $n_0inmathbb{N}$ such that $2^{n_0-1}<k<2^{n_0}implies dfrac{1}{2^{n_0}}<dfrac{1}{k}<dfrac{1}{2^{n_0-1}}$. My claim is that $xin left(dfrac{1}{2^{n_0}},dfrac{3}{2^{n_0}}right)$, and approach was to try and prove that $left(dfrac{1}{k},dfrac{1}{k-1}right)subseteq left(dfrac{1}{2^{n_0}},dfrac{3}{2^{n_0}}right)$, but I am unable to proceed ahead from this point.

I may be wrong as well, of course. Any help would be highly appreciated, in either proving or disproving the claim that I make in the last sentence.

general topology – Proving the equivalence of some different definitions of continuity

I wanted to verify the correctness of my solution to a basic point-set topology exercise. Also, could the ($2 leftrightarrow 3$) component of the proof be made more direct and enlightening?

Proposition. Let $f : X rightarrow Y$ be a function from space $X$ to space $Y$. The following conditions are equivalent:

  1. given any open $U subseteq Y$, $f^{-1}(U)$ is open,
  2. given any closed $F subseteq Y$, $f^{-1}(F)$ is closed, and
  3. given any subset $A subseteq X$, $f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$.

Proof.

Let $f : X rightarrow Y$ be a function from space $X$ to space $Y$.

(1 $leftrightarrow$ 2) $f^{-1}(U)$ is open for every open $U subseteq Y$ $leftrightarrow$ $f^{-1}(U)^c$ is closed for every open $U subseteq Y$ $leftrightarrow$ $f^{-1}(U^c)$ is closed for every open $U subseteq Y$ $leftrightarrow$ $f^{-1}(U^c)$ is closed for every $U^c subseteq Y$ where $U$ is an open set of $Y$ $leftrightarrow$ $f^{-1}(F)$ is closed for every closed $F subseteq Y$ since complements of opens are exactly the closed sets.

(2 $rightarrow$ 3) Suppose $f^{-1}(F)$ is closed for every closed subset $F subseteq Y$. We need to show that $f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$ for every subset $A subseteq X$. Let $A subseteq X$ be an arbitrary subset of $X$ and let $x in mathsf{Clos}(A)$. It remains to be shown that $f(x) in mathsf{Clos}(f(A))$ i.e. $x in f^{-1}(mathsf{Clos}(f(A)))$. By assumption, we have that $f^{-1}(mathsf{Clos}(f(A)))$ is closed so it suffices to show that $x in mathsf{Clos}(f^{-1}(mathsf{Clos}(f(A))))$. We are done if we can show that every open neighbourhood of $x$ intersects $f^{-1}(mathsf{Clos}(f(A)))$. Let $U$ be an arbitrary open neighbourhood of $x$. Because $x in mathsf{Clos}(A)$, $U$ must contain some $a in A$. Clearly, $f(a) in f(A) subseteq mathsf{Clos}(f(A))$ which is to say that $a in f^{-1}(mathsf{Clos}(A))$.

(3 $rightarrow$ 2) Suppose $f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$ for every subset $A subseteq X$. Let $F subseteq Y$ be a closed subset. We need to show that $f^{-1}(F)$ is closed i.e. $mathsf{Clos}(f^{-1}(F)) subseteq f^{-1}(F)$ which is equal to $f^{-1}(mathsf{Clos}(F))$ because $F$ is closed. Let $x in mathsf{Clos}(f^{-1}(F))$. We have by assumption that
begin{equation*}
f(mathsf{Clos}(f^{-1}(F))) subseteq mathsf{Clos}(f(f^{-1}(F))) = mathsf{Clos}(F)
end{equation*}

meaning $f(x) in mathsf{Clos}(F)$ i.e. $x in f^{-1}(mathsf{Clos}(F))$ which was what we needed.

Proving cardinality given surjection from A to B and B to A

Suppose f:A→B and g:B→A are both surjective, does this imply that there is a bijection between A and B.

I was told that the statement above is true only with Axiom of Choice. Can someone provide an example of why that is the case?