## complexity theory – Padding in proof of space hierarchy theorems

Suppose that we consider instead the language
$$L = { langle M rangle : text{M does not accept langle M rangle in space f(langle M rangle)} }.$$
We want to show that $$L notin mathsf{SPACE}(o(f(n))$$, that is, that if $$M$$ uses space $$o(f(n))$$ then $$L(M) neq L$$. This should be the case since $$langle M rangle in L Leftrightarrow langle M rangle notin L(M).$$
But is this really true? According to the definition of $$L$$, $$langle M rangle in L$$ iff $$M$$ does not accept $$langle M rangle$$ in space $$f(langle M rangle)$$. It could be that $$langle M rangle in L$$ and $$M$$ accepts $$langle M rangle$$ using more than $$f(langle M rangle)$$ space. The latter could actually happen, since we are only guaranteed that $$M$$ uses space $$g(n)$$ for some function $$g(n) = o(f(n))$$, which does not preclude $$g(|langle M rangle|) > f(|langle M rangle|)$$ at the particular value $$|langle M rangle|$$.

Adding the padding fixes this issue: it cannot be that $$g(|(langle M rangle, 10^k)|) > f(|(langle M rangle, 10^k)|)$$ for all $$k$$, since this would contradict $$g(n) = o(f(n))$$.

## real analysis – Wrong inequality in simple proof Cauchy \$implies\$ Boundedness.

These notes from Oxford University contain an apparently very simple proof that Cauchy sequences (real or complex) imply boundedness.

I understand the Cauchy condition $$|a_m – a_n| < epsilon$$, and that the proof assigns $$epsilon=1$$ as an arbitrary value.

Question: I can’t understand how the following inequality is derived from the triangle inequality:

$$|a_m| leq 1 + |a_N|$$

My Attempt: I have tried using the reverse triangle inequality with no success:

$$|a_m| – |a_N| leq |a_n -a_N| < 1$$

And so,

$$|a_m| < 1 + |a_N|$$

Here the inequality is $$<$$ and not $$leq$$ as per the reproduced notes.

For convenience, the proof is reproduced below.

## probability theory – Donsker and Varadhan inequality proof without absolute continuity assumption

I’ve been attempting to understand the proof of the Donsker-Varadhan dual form of the Kullback-Liebler divergence, as defined by
$$operatorname{KL}(mu | lambda) = begin{cases} int_X logleft(frac{dmu}{dlambda}right) , dmu, & text{if mu ll lambda and logleft(frac{dmu}{dlambda}right) in L^1(mu),} \ infty, & text{otherwise.} end{cases}$$
$$operatorname{KL}(mu | lambda) = sup_{Phi in mathcal{C}} left(int_X Phi , dmu – logint_X exp(Phi) , dlambdaright).$$

Many of the steps in the proof are helpfully outlined here: Reconciling Donsker-Varadhan definition of KL divergence with the “usual” definition, and I can follow along readily.

However, a crucial first step is establishing that (for any function $$Phi$$)
$$tag{1}label{ineq} operatorname{KL}(mu|lambda)ge left{int Phi dmu-logint e^{Phi}dlambdaright},$$
said to be an immediate consequence of Jensen’s inequality. I can prove this easily in the case when $$mu ll lambda$$ and $$lambda ll mu$$:

$$operatorname{KL}(mu|lambda) – int Phi dmu = int left( -logleft(frac{e^{Phi}}{dmu / dlambda}right) right) dmu ge -log int frac{e^{Phi}}{dmu / dlambda} dmu = -logintexp(Phi)dlambda.$$
However, this last step appears to crucially rely on the existence of $$dlambda/dmu$$ and thus that $$lambda ll mu$$, which isn’t assumed by the overall theorem. Where I have been able to find proofs of the above in the machine learning literature, this assumption seems to be implicitly made, but I don’t believe it is necessary and it is very restrictive.

My question is: how can we prove ref{ineq} without assuming $$lambda ll mu$$?

## proof writing – If \$X\$ is compact then prove that \$X\$ is complete and totally bounded.

I tried to do it in a way different from my textbook:

Let $$(X,d)$$ be a compact metric space then it is totally bounded.(I have been Ble to prove this).

My doubt lies in the following part that I have tried to prove:

Let $$X$$ be a compact metric space then $$X$$ is totally bounded.We choose a cauchy sequence $${x_n}$$ in $$(X,d)$$ .

Let $$A$$ be a subset of $$(X,d)$$ such that $$A={x_n}$$.Then $$A$$ will be totally bounded too as $$A subset (X,d)$$.

Let for an $$epsilon > 0$$ there exist points $${x_1′,cdots x_n’}$$ such that $$B_d(x_k’,epsilon)$$ contains infinitely many points of $$A$$.Then as $${x_n}$$ is a cauchy sequence so we can conlcude that $${x_n}$$ converges to $$x_k’$$. Also as $$(X,d)$$ is a metric space so $${x_n}$$ can converge to only one points and we have shown that $$x_k’$$ is a limit point of $${x_n}$$.

## Another proof for interior of \${(x,y) mid y=0} cup {(x,y) mid x>0 text{and} y neq 0}\$

Consider the set $$C= A cup B ={(x,y) mid y=0} cup {(x,y) mid x>0 text{and} y neq 0}$$ I want to prove $$text{Int } B = text{Int } C$$.
My attempt:

Clearly $$text{Int } B subset text{Int } C$$ as $$B subset C$$. To prove the converse inclusion, let $$(x,y) in text{Int } C$$, then there must exists a $$r>0$$ such that $$B_d((x,y), r) subset C = A cup B$$.

Actually I want to show that $$B_d((x,y), r) cap A = emptyset$$, which implies that $$B_d((x,y), r) subset B$$.

## functions – Help with Strong Induction proof

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## real analysis – Help understanding the Proof of Cauchy’s criterior for uniform convergence of sequence of functions

Before asking the question I would like to say that I filtered it into mathstackexchange before, and that every answer to a similar question didn’t really convinced me, so please don’t just write “go see this answer” and leave. That being said, my question is: i’ve seen two types of proofs for this theorem: the one that uses a limit argument at a certain point, for which I don’t understand why are we allowed to use that, since the limit in question is a pointwise limit, and so one needs to fix X in order to make it work, but then claims that the obtained inequality holds for every X in the set for which the sequence {f_n(X)} is uniformly Cauchy. Why can this be done? To me it feels like it’s not that rigourous. Is there any other epsilon-N proof which I can understand (I saw a couple of them on this site but I couldn’t figure out why they worked, since they have a similar flaw of the proof above: they use an argument that to me is justified only when X is fixed, while the X is clearly not fixed. Or, if they fix it, then they claim that it works for every X on the set for which f_n is uniformly Cauchy)

## proof techniques – Mutual Friends in a Network?

I always seem to have trouble finding a formal way to analyze this (be through proofs or whatever).

The problem statement is as such:

If A and B are friends, and B and C are friends, then A and C are friends too.

In a simple network like the following, this makes complete sense:

1 — 2 — 3

Analysis:
We can see that 1 and 2 are friends, and 2 and 3 are friends. It follows from the problem statement that 1 and 3 must be friends too. This is the most generic case for a problem like this.

Where I get confused is in a following network:

1 — 2 — 3 — 4

Analysis:
We can see that 1 and 2 are friends, and 2 and 3 are friends; therefore, 1 and 3 must be friends. Also, since 2 and 3 are friends, and 3 and 4 are friends; therefore, 2 and 4 must also be friends.

Since 1 and 2 are already friends, would it follow from our conclusion of the last sentence (that 2 and 4 are friends) that 1 and 4 must also be friends?

Moving forward into a bigger picture, any group of connected nodes would also all be friends?

What’s the best way to analyze this?

## computer science – Pumping Lemma proof for this language: L={a^ib^jc^k∣k=i∗j}

i have troubles to show that this language is not context free with the pumping lemma.

As a word I chose: a^mb^mc^(m^2)

I solved all the cases but one, which is:
“vxy contains b’s and c’s”

I came up with the following approach to show that the resulting word can not be in the language:

p * (p+|v|) != p*p + |y|

this is equal to:

p * |v| != |y|

Because p must >= 1 and |vxy| > 0 and |vxy| <= p the resulting word is not in the language.
Is this assumption correct?

## Proof of limit of \$A(P_n) – A(p_n) = 0\$ for exhaustion method

This is the last question on an exhaustion method problem set.

I need to show
$$lim_{n rightarrow infty} n(sin(frac{2pi}{n}) – tan(frac{2pi}{n}) ) = 0$$

Without using L’ Hospital’s rule…
Any pointers on where to start?