Continuous injective map $f:\mathbb{R}^3 \to \mathbb{R}$? How would you show that there is no continuous injective map  $f:\mathbb{R}^3 \to \mathbb{R}$?
I tried the approach where if $a \in \mathbb{R}$ then $\mathbb{R} \backslash \{a\} $ is not clearly connected so there can't exist a continuous function otherwise $\mathbb{R} \backslash \{a\} $ would be connected?
 A: I will present an answer which can be (in principle, at least) understood by anyone who knows single variable calculus and the definition of a continuous function $f: \mathbb{R}^n \rightarrow \mathbb{R}$.  Then I will explain how to shorten the argument a little by using topological language.
Step 1: Let $f: [0,1] \rightarrow \mathbb{R}$ be a continuous function with $f(0) = f(1)$.  Then there exist $x,y \in [0,1)$ such that $f(x) = f(y)$ and $x\neq y$.  
Proof: We may assume $f$ is nonconstant.  By the Extreme Value Theorem it assumes a minimum value $m$ and a maximum value $M$ with $m < M$.  Let $x_m,x_M$ be such that 
$f(x_m) = m$ and $f(x_M) = M$.  Without loss of generality $x_m < x_M$. By the Intermediate Value Theorem, every value in $(m,M)$ is assumed on the interval $(x_m,x_M)$.  Moreover, because $f(1) = f(0)$, the function 
$g: [x_M,1+x_m]: \rightarrow \mathbb{R}$ given by 
$x \mapsto f(x)$, $x_M \leq x \leq 1$,
$x \mapsto f(x-1)$, $1 \leq x \leq 1+x_m$
is continuous, with $g(x_M) = M$, $g(1+x_m) = m$, so by the Intermediate Value Theorem takes every value in $(m,M)$ on the interval $(x_M,1+x_m)$, so that $f$ takes every value in $(m,M)$ on $(x_M,1) \cup (0,x_m) = [0,1] \setminus [x_m,x_M]$.  Thus $f$ 
takes every value in $(m,M)$ at least twice and is not injective on $[0,1)$.  
Step 2: Let $n$ be an integer greater than one, and let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous function.  Then $g: [0,1] \rightarrow \mathbb{R}$ given by 
$g(t) = f(\cos(2\pi t),\sin(2\pi t),0,\ldots,0)$ is continuous with $g(0) = g(1)$, so by Step 1 there is $0 \leq t_0 < t_1 < 1$ such that $g(t_1) = g(t_2)$.  That is,
$f(\cos(2\pi t_1),\sin(2\pi t_1),0,\ldots,0) =  f(\cos(2\pi t_2),\sin(2\pi t_2),0,\ldots,0)$, so $f$ is not injective.
Step 3: A softer, more topological version of this is as follows: let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be continuous.  Seeking a contradiction, we suppose it is injective. Let $S^{n-1} \subset \mathbb{R^n}$ be the unit sphere.  Since it is compact, the restriction of $f$ 
to $S^{n-1}$ gives a homeomorphism onto its image, which is a compact, connected subset of $\mathbb{R}$ hence a closed bounded interval $[a,b]$.  If $a = b$ then $f$ is constant, hence not injective.  Otherwise, observe that if we remove any one of the uncountably infinitely many points from $S^{n-1}$ we get a space homeomorphic to $\mathbb{R}^{n-1}$, which is connected if $n \geq 2$.  However, there are only two points in $[a,b]$ whose removal leads to a connected space: the two endpoints.  Contradiction!
A: Another version that requires a lot less of topology, but only the intermediate value theorem is the following.
Let $f \colon \mathbb{R}^n \to \mathbb{R}$ be continuous, with $n \geq 2$. Without loss of generality, we may assume $f(0,\ldots,0) = 0$.
Consider the curve $\gamma_1 \colon [-1,1] \to \mathbb{R}^n; x \mapsto (x,0,\ldots,0)$, and consider $g := f \circ \gamma_1$. If both $a_1 := g(-1)$ and $a_2:= g(1)$ have the same sign, proceed as follows. Without loss of generality, assume $a_1 \leq a_2$. Apply the intermediate value theorem to find a $c \in [0,1]$ with $g(c) = a_1$. Then $g(-1) = g(c)$, with $c \ne -1$. Hence $f(\gamma(-1) = f(\gamma(c))$ (and $\gamma(-1) \ne \gamma(c)$).
Consider on the other hand that $a_1$ and $a_2$ have different signs, consider the curve $\gamma_2 \colon [-1,1] \to \mathbb{R}^n; x \mapsto (0,x,\ldots,0)$ and $h(x) = f(\gamma_2(x))$. Obviously at least two of $h(-1), h(1), g(-1)$ and $g(1)$ will have the same sign. Apply the earlier reasoning to find two points $c$ and $d$ with $f(c) = f(d)$.
This obviously extends to show that no continuous $f \colon U \to \mathbb{R}$ with $U$ open in $\mathbb{R}^n$ can be injective (for $n \geq 2$).
A: Take two lines through the origin $\ell_1=tb_1,\ell_2=tb_2$ where $b_1,b_2 \in \Bbb{R}^3$. Then $g_i(t)=f(tb_i)$ are injective and continuous, and therefore strictly monotone. This proves that $f(\ell_1)\cap f(\ell_2)$ cannot equal a single point, therefore contradicting injectivity.

Another approach. Take $a,b \in \Bbb{R}^3$ such that $f(a)<f(b)$. Then the image of every bounded connected path between $a$ and $b$ must be the interval $[f(a),f(b)]$. This contradicts injectivity.
A: We can prove there is no continuous injection from $\mathbb{R}^{2}$ to 
 $\mathbb{R}.$Then we confine the map from $\mathbb{R}^{3}$ to $\mathbb{R}$ on the $xy$ plane.The map is still continuous and injective so we have contradction.
We just need to find 2 different points and $f$ has same value on the two points.
So just consider 4 vertexs $\{O(0,0),A(1,0),B(0,1),C(1,1)\}$ of the unit cube.We can assume $f$ takes different value on the $4$ points or we just find what we want.Assume $f(O)$ is the smallest and $f(O)<f(A)<f(B)$.Using mean value theorem and we can find a point $D$ on the segment $OB$ and $f(D)= f(A).$
This method just needs continuity for each variable and can be generalized to the case of $\mathbb{R}^{n}.$
A: Suppose such an $f$ exists. Then, since $f$ is continuous, $f(\Bbb R^3)$ is a connected subset of $\Bbb R$ and hence, an interval $U$.
Also, since $f$ is one-to-one, the  following two properties hold: 
$\ \ $1) $U$ is not a singleton point, 
and, 
$\ \ $2) for each $a\in\Bbb R^3$, we have $f\bigl(\, \Bbb R^3\setminus\{a\}\,\bigr) = f(\Bbb R^3) \setminus \{\,f(a)\,\} =U\setminus\{\,f(a)\,\}$.
Now, by 1), we may (and do) choose $b\in\Bbb R^3$ such that $f(b)\in\text{Int}(U)$, where $\text{Int}(U)$ is the interior of $U$.
Then, by 2), 
$$
  f\bigl(\, \Bbb R^3\setminus\{\,b\,\}\,\bigr) = U \setminus \{\,f(b)\,\};
$$
which is a contradiction, since $\Bbb R^3\setminus\{\,b\,\}$ is connected whilst $U \setminus \{\,f(b)\,\}$ is not.
A: Suppose there exist continuous injective map $f: \mathbb R^3 \to \mathbb R$ is continuous injective , then 
$f:\mathbb R^3 \to f(\mathbb R^3)$ is continuous bijective , then $f(\mathbb R^3)$ is an infinite connected set so a non-singleton 
interval in $\mathbb R$ say $I:= f(\mathbb R^3)$ , but then for any  three distinct $a,b,c \in \mathbb R^3$ , $f(a),f(b),f(c)$ are three 
distinct elements of $I$ and $f:\mathbb R^3\setminus \{a,b,c\}  \to I\setminus \{f(a),f(b),f(c)\}$ is a continuous bijection , so 
that $I\setminus \{f(a),f(b),f(c)\}$ must be connected i.e. an interval of $\mathbb R$ , but it is known that for any 
non-singleton interval $I$ in $\mathbb R$ and any three distinct points $x,y,z \in I$ , the set $I \setminus \{x,y,z\}$ cannot 
be an interval , contradiction !
[NOTE : Let $I \subseteq \mathbb R$ be a non-singleton interval , $x,y,z \in I$ be three distinct points , w.l.o.g. 
$x<y<z$ , then as I is an interval , $\dfrac {x+y}2 , \dfrac {y+z}2 \in I$ and also $x<\dfrac{x+y}2<y<\dfrac {y+z}2<z$ , so 
that $\dfrac {x+y}2 , \dfrac {y+z}2 \in I\setminus \{x,y,z\}$ , then if $I\setminus \{x,y,z\}$ were an interval , the fact
$\dfrac {x+y}2 <y< \dfrac {y+z}2$ would imply $y \in I\setminus \{x,y,z\}$ , contradiction ! Thus $I\setminus \{x,y,z\}$  cannot be an 
interval ]
