# Tag Info

Accepted

### Showing permutation does not change output of commutative operation (recursion theorem)

First of all, I don't believe you have sufficient foundation in basic logic to tackle such problems. This is not your fault, naturally, because it is hard to find good teachers. For you to truly grasp ...

### Why is the Cantor Set not a subset of $\mathbb{Q}$?

Let's start with what you know! So, you know that there is a bijection between all ternary strings with only zeroes and twos and the cantor set. In the proof, what is used is the fact that the Cantor ...
Accepted

### Is this an injection from $\mathbb{R}_+$ to $(0,1)$?

Your proposal is rather unclear, but per your comments it does not give a function with codomain $\mathbb{R}$ at all; this is because things like [a sequence] of the form 0.0000000000…33333333… where ...
Accepted

### Prove that $d(X,Y) = |X\setminus Y| + |Y\setminus X|$ is a distance

Hint: If $X^c$ denotes the complement of $X$ then: $|X\setminus Z|=|X\cap Z^c|=|X\cap Z^c\cap Y|+|X\cap Z^c\cap Y^c|.$ And you can similarly decompose $|Z\setminus X|$. Can you find an upper bound for ...
Accepted

### Is there reason for concern with this proof that "there is a bijective function from $\{1,...,m\}$ to $\{1,...,n\}$ only if $m = n$"?

In the context of proving a statement like this, you are correct that you should really never be using the notation "Card" as if it were a function. But the proof still works when you ...

### Confused about the measure of the Cantor Set, and how to reconcile this with there being points not at the endpoints

so eventually we will get $\lim_{n\rightarrow \infty} (\frac 2 3)^n = 0$ left, such that nothing remains except the endpoints. This sentence is where your intuition has led you astray, and led you to ...
1 vote

### How can it be that the empty set is a subset of every set but not an element of every set?

Element: An element x is said to be an element of a set A if x is contained within the set A. This is denoted as: x ∈ A. Subset: A set A is considered a subset of another set B if every element of ...
1 vote
Accepted

### Prove that $C(X \times Y) = A \times C(Y) \cup C(X) \times B$

I think it is implicit that: the complement of $X$ is meant to be with respect to $A$ the complement of $Y$ is meant to be with respect to $B$ the complement of $X \times Y$ is meant to be with ...
1 vote
Accepted