I am trying to find a 5-dimensional injective function, any suggestions or any ideas as to how I can construct one? Formally I am looking for a function $f: \mathbb{R}^5 \rightarrow \mathbb{R}$ which is injective.
Constructing a 5-dimensional injective function
functions
Related Solutions
First note that $\mathbb{W}$ is just shifted $\mathbb{V}$ so regarding the question, there is no difference in those four cases.
1) There exist an injective mapping from $\mathbb{V} × \mathbb{V}$ into $\mathbb{R}$. Since $\mathbb{V} ⊆ \mathbb{R}$, we have a nice injective mapping, even an embedding $\mathbb{V} × \mathbb{V} \to \mathbb{R} × \mathbb{R}$. And there is a theorem stating that for every infinite set $A$, there is a bijection between $A$ and $A × A$. One can imagine the base case of the theorem – enumerating pairs of natural numbers (starting with the origin and concatenating the finite diagonal lines in the first quadrant). So by composing our embedding with a bijection between $\mathbb{R} × \mathbb{R}$ and $\mathbb{R}$, we get desired injective mapping.
2) As said in comments, there is no continuous injective function $\mathbb{V} × \mathbb{V} \to \mathbb{R}$, because the continuous injectivity could then be promoted to embedding on any compact subspace, so we would get a two-dimensional compact subspace of one-dimensional real-line, which is not possible.
3) One can construct an explicit injective function $\mathbb{V} × \mathbb{V} \to \mathbb{R}$. Real numbers are almost the same thing as infinite sequences of ones and zeroes. Informally, you can imagine taking a decimal expansion of the real, but using binary digits instead, the only problem is that there are two different expansions of each rational number – the one ending with $1000…$ and the one ending with $0111…$. Formally, there is a quotient function $2^ω \to [0, 1]$. $2^ω$ is a set of all functions from $ω$ to $2 = \{0, 1\}$, i.e. the set of all infinite sequences of zeroes and ones. It has also a structure of a topological space – taking the product topology of infinite product of $2$-point discrete spaces. This topological space is called Cantor space and is homeomorphic to Cantor set in reals (http://en.wikipedia.org/wiki/Cantor_set). The quotiet just glues the endpoints of the gaps in the Cantor set (i.e. the two equivalent binary expansions of rational numbers) together. So any irrational number has one preimage and any rational number has two preimages. By choosing one preimage for each irrational number, you get an injective function $[0, 1] \to 2^ω$ which is not continuous, but is kind of nice, as created from that quotient mapping. So now we have $[0, 1] × [0, 1] \to 2^ω × 2^ω \to 2^ω \to [0, 1] ⊆ \mathbb{R}$, since $2^ω × 2^ω$ is homeomorphic to $2^ω$ – from a pair of sequences you form one sequence by intertwining the sequences – you realize first sequence on odd indices and the second on even indices.
Yes, there is such a function. Define $f: \Bbb Z \times \Bbb Z \to \Bbb Z$ by $$f(m,n) = \begin{cases} 2^{m}3^{n} & m, n \geq 0 \\ 5^{-m}7^{-n} & m, n < 0 \\ 11^{m}13^{-n} & m > 0, n < 0 \\ 17^{-m}19^{n} & m < 0, n > 0 \end{cases}.$$ You should prove this is injective. For it not being surjective: which pair of numbers is being sent to any negative integer? The range of $f$ is a subset of the positive integers.
Best Answer
First we construct an injective function $\mathbb{R}^2\rightarrow \mathbb{R}.$ To simplify matters, it suffices to construct an injective function $(0,1)\rightarrow \mathbb{R}$ since $\tan(\pi (x-\frac{1}{2}))$ is a bijection $(0,1)\rightarrow \mathbb{R}.$
Given a pair of real numbers $(a,b)\in(0,1)^2$ they have unique binary representations $a = \sum_{i=0}^{\infty}\frac{a_i}{2^i}$ and $b = \sum_{i=0}^{\infty}\frac{b_i}{2^i}$ with neither $(a_i)$ nor $(b_i)$ ending in an infinite tail of $1$'s. Then it's not hard to see that the map sending $(a,b)$ to $0.a_1 b_1 a_2 b_2 \ldots$ (written in binary) is an injective map $(0,1)^2\rightarrow (0,1).$ Notice that this decimal will not end in infinite sequence of $1$'s by construction.
Now by identifying $\mathbb{R}^5$ with $\mathbb{R}^2\times \mathbb{R}^3,$ we obtain an injective map $\mathbb{R}^5\rightarrow \mathbb{R}\times \mathbb{R}^3 = \mathbb{R^4}.$ Similarly, we have injective maps $\mathbb{R}^4\rightarrow \mathbb{R}^3$, $\mathbb{R}^3\rightarrow \mathbb{R}^2$, and $\mathbb{R}^2\rightarrow \mathbb{R}.$ By composing these maps we get an injection $\mathbb{R}^5\rightarrow \mathbb{R}$ as desired.