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Apr
23
comment Are we allowed to compare infinities?
than another then we can view it as 'larger' in some sense.
Apr
23
comment Are we allowed to compare infinities?
@SimpleArt Yes it does. The point is that there are many different ways we can compare two infinite sets. The OP had an intuition that there are more multiples of $2$ than multiples of $3$, and it's important to see how that intuition has a mathematical foundation. Similarly, she might feel that there are many more whole numbers than prime numbers. When talking about density, I suppose it is important to stress that if two sets have the same density then they are not necessarily the same size - so there are not $0$ prime numbers - but that if one set has a strictly higher density
Apr
22
comment Are we allowed to compare infinities?
If you do 'think about stopping', then you arrive at a useful notion - that of density. The multiples of $2$ have density $1/2$ in the natural numbers, while the multiples of $3$ have density $1/3$, so in that sense the OP's intuition that there are more multiples of $2$ than multiples of $3$ is justified.
Apr
18
comment Product topology is discrete
That is right. $\quad$
Apr
17
comment Product topology is discrete
@HennoBrandsma I would too, but it turned out to be remarkably difficult to write the sentence that way without the order of quantifiers being ambiguous. I completely agree that it's not ideal.
Apr
17
comment Product topology is discrete
@Silent The reverse implication in my statement is that if $\Pi X_\alpha$ is discrete for all collections $X_\alpha$ of discrete spaces, then $A$ is finite. If $A$ is infinite, there may be particular cases in which $\Pi X_\alpha$ is discrete - such as when $X_\alpha=\{1\}$ for all $\alpha$, but there must be at least one collection of discrete $X_\alpha$ such that $\Pi X_\alpha$ is not discrete. Henno's answer gives a stronger statement - the case when $\Pi X_\alpha$ is discrete is precisely the case when all but finitely many $X_\alpha$ have one point.
Apr
17
answered Product topology is discrete
Apr
9
comment existence of a lifting
For a suitably well behaved space $Y$, the failure of this to hold in general is captured by the first (co)homology group of $Y$. Loosely speaking, every singular simplex within $Y$ may be realized as a homotopy and so admits liftings by the homotopy lifting property. But a cycle (simplex with no boundary) within $Y$ admits no lift in general, as in the example for $S^1$ given by Tsemo Aristide.
Apr
8
comment Understanding Euclid's proof that the number of primes is infinite.
@MichaelHardy An answer to this question does not require a diatribe into the historical background of Euclid's proof or whether or not it requires contradiction. There are two ways to answer the question: 1) Point out that even if $1+p_1\dots p_n$ is not prime, it must have some prime divisor that is not one of the $p_n$ or 2) Claim that the entire proof is by contradiction, so there's no sense trying to derive anything from it. I agree with you that (1) is far better, but it is sufficient to give the argument without claiming that approach (2) is erroneous.
Apr
8
comment Understanding Euclid's proof that the number of primes is infinite.
@MichaelHardy (math.stackexchange.com/questions/631977/…) I understand this is a pet peeve for you, but maybe it's time to let this one go. This proof is perfectly valid - maybe not as nice as yours, and maybe not what Euclid wrote, but that's by the by. Note that the OP was asking specifically for a proof that the number of primes is infinite, not for a proof that if $S$ is any finite set of primes, then there's a prime not contained in $S$.
Apr
8
comment Show that $H_n$ is almost always a non-terminating decimal.
I've edited my answer to make that point more clear.
Apr
8
revised Show that $H_n$ is almost always a non-terminating decimal.
added 45 characters in body
Apr
8
comment Show that $H_n$ is almost always a non-terminating decimal.
The first condition is what we assume, which allows us to deduce things about $p$. The second one is what we need to use in order to apply the induction hypothesis. Note that if $\frac12((k+1)+1)\le p\le k$ then it is certainly true that $\frac12(k+1)\le p\le k$. That doesn't mean that $p$ could be equal to $\frac12(k+1)$.
Apr
8
comment Show that $H_n$ is almost always a non-terminating decimal.
@AgnetaGauck I think you need to read my proof again. I don't say that those two conditions are the same. Instead, I am assuming that the first condition is true. If that is the case, and if it is also true that $p<k+1$ then the second condition is true and we can apply the inductive hypothesis. There is no point in the proof at which $p$ could be equal to $\frac12(k+1)$.
Apr
8
revised Show that $H_n$ is almost always a non-terminating decimal.
tags
Apr
8
answered Show that $H_n$ is almost always a non-terminating decimal.
Apr
7
comment A question about lonely runner conjecture
OK. You're correct.
Apr
7
revised A question about lonely runner conjecture
added 1055 characters in body
Apr
7
answered A question about lonely runner conjecture
Apr
6
revised Max possible line of sight on earth
edited tags