Does there exist a continuous bijective function $ \displaystyle f : \Bbb R \rightarrow \Bbb R − \{1\}$? This is my first time posting on here :)
My question is this: Does there exist a continuous bijective function $f :\Bbb R → \Bbb R − \{1\}$? Explain yes or no...
My thoughts:
Let $ \displaystyle f:  A \rightarrow  B$ be a function 
Okay for there to be a bijective function, for every element in A, there must be a unique element mapping $A$ to $B$. Right? Well if that is the case, then the answer to the question is FALSE. Why? Because the cadinality of the domain of $A$ (which in this case is R) is GREATER than the cardinality of the co-domain which is ($\Bbb R-\{1\}$). 
Because this is so, there must be some duplicates such that two different elements in $A$ map to the same element in $B$.
Does this make sense?
Unfortunately, I would be happy with this answer but another question pops up in my mind. I have read in my textbook that the cardinality of $\Bbb N$ is equal to the cardinality of $\mathbb Z$  
But how? Since $\Bbb N$ represents all natural numbers ($0,1,2,3...$) and $\mathbb Z$ represents all the integers ($...-3,-2,-1,0,1,2,3...$). Clearly the cardinality of $\mathbb Z$ is greater than N but still the two have equal cardinality. 
So tying that in with the question, my answer doesn't seem right anymore.
Any guidence or help would be greatly appreciated!!!
Thanks :)
 A: You can't do this by cardinality arguments, since $\mathbb{R}$ and $\mathbb{R}\setminus\{1\}$ actually have the same cardinality. You need some topological argument.
The keyword is connectedness. The continuous image of a connected set like $\mathbb{R}$ must be connected, unlike $\mathbb{R}\setminus\{1\}$. So your function can't even be continuous and onto. 
Even though it boils down to the same in $\mathbb{R}$, it is slightly easier to consider convexity. Recall a set $S$ is convex in an affine space if, for every $s_1,s_2$ in $S$, and every $t\in [0,1]$, the element $(1-t)s_1+ts_2$ belongs to $S$. That is, $[s_1,s_2]\subseteq S$ as soon as $s_1,s_2$ belong to $S$.
By the intermediate value theorem, the continuous image of a convex set in $\mathbb{R}$ is convex. Now $\mathbb{R}$ is convex, but $\mathbb{R}\setminus\{1\}$ is not. So there does not even exist a continuous function from $\mathbb{R}$ onto $\mathbb{R}\setminus\{1\}$.
A: The concept of cardinality does not consider any particular structure imposed on a set (though they may help us count). In other words, it is the extra structure (such as operations or order relations) that misled you to a wrong claim that $\Bbb{Z}$ has more element than $\Bbb{N}$. Only when you obliterate this extraneous information you can truly compare the cardinality of two sets.
For example, let us rename each integers as follows:
$$ \Bbb{Z} = (0, 1, -1, 2, -2, 3, -3, \cdots) = (a_1, a_2, a_3, a_4, a_5, a_6, a_6, \cdots). $$
In this way we can forget everything about the size, order and arithmetic relationship of the elements of $\Bbb{Z}$. Then now we identify the function $n \mapsto a_n$ as a bijection between $\Bbb{Z}$ and $\Bbb{N}$, hence conclude that they have the same cardinality.
A: The cardinalities of $\Bbb R$ and $\Bbb R\setminus\{1\}$ are actually the same, even though it may seem that they shouldn't be. It's clear that $\Bbb R$ is "at least as big" as $\Bbb R\setminus\{1\}$, since the former contains the latter. On the other hand, the map $\Bbb R\to\Bbb R\setminus\{1\}$ given by $x\mapsto1+e^x$ is a one-to-one function, so $\Bbb R\setminus\{1\}$ is "at least as big" as $\Bbb R$. Thus, by this theorem, the two sets are "the same size".
The kicker here is that a continuous real-valued function on $\Bbb R$ can only have one of the following types of sets as its range:

(a) A singleton $\{y\}$--this happens for constant functions.
(b) An interval (half-open, open, or closed).
(c) A ray (open or closed).
(d) The whole real line.

$\Bbb R\setminus\{1\}$ isn't any of these types of sets, so no continuous bijection $\Bbb R\to\Bbb R\setminus\{1\}$ exists.

Alternately, we could use the Intermediate Value Theorem, which can also be used to prove the "kicker" mentioned above.
Let $f$ be any continuous function $\Bbb R\to\Bbb R\setminus\{1\}$. If $f$ takes on at least one value greater than $1$ and at least one value less than $1$, then there exist distinct $a,b$ such that $f(a)<1<f(b)$. But since $f$ is continuous on the interval $[a,b]$ or $[b,a]$ (whichever one makes sense), then by IVT, there would then be some $c$ strictly between $a$ and $b$ such that $f(c)=1$, which is impossible, since $f:\Bbb R\to\Bbb R\setminus\{1\}$. Thus, for any continuous function $f:\Bbb R\to\Bbb R\setminus\{1\}$ either $f(x)>1$ for all $x$ or $f(x)<1$ for all $x$, so $f$ is not a bijection $\Bbb R\to\Bbb R\setminus\{1\}$. Thus, no continuous bijection $\Bbb R\to\Bbb R\setminus\{1\}$ exists.
A: No. Because $\mathbb R$ is connected but $\mathbb R-\{1\}$ is not connected.
