Prove that there does not exist an $1-1$ and continuous mapping from $\mathbb{R}^2$ to $\mathbb{R}$.
Solution
In general there does not exist an $1-1$ and continuous mapping from $\mathbb{R}^n$ to $\mathbb{R}$ for $n \geq 2$. The first proof we are giving is an easy one. The second one is a bit more technical.
1st proof: Due to connectness we have that $f \left( \mathbb{R}^2 \right)= \mathcal{I}$ where $\mathcal{I}$ is an interval. Note that if we remove a point from the plane, it still remains connected. Having that in mind we observe that $f \left ( \mathbb{R}^2 \setminus f^{-1}\left ( a \right ) \right )$ is connected and equals $\mathcal{I} \setminus \left \{ a \right \}$ whereas this is not connected leading to a contradiction.
2nd proof: Define $\mathcal{A}_x = f(\{x\}\times\mathbb {R})$ for each $x \in\mathbb {R}$. Since $\{x\}\times\mathbb {R}$ is connected and $f$ is continuous, $\mathcal{A}_x$ must also be connected. Therefore it is an interval. It is not degenerate (a point), because $f$ is injective. Therefore each $\mathcal{A}_x$ contains an open interval $(\mathfrak{p}_x, \mathfrak{q}_x)$, where $\mathfrak{p}_x$ and $\mathfrak{q}_x$ are distinct rational numbers. Since there are only countably many such intervals $(\mathfrak{p}_x, \mathfrak{q}_x)$ and uncountably many sets $\mathcal{A}_x$, there exist $x \neq y$ such that $(\mathfrak{p}_x, \mathfrak{q}_x)=(\mathfrak{p}_y, \mathfrak{q}_y)$. Thus the sets $\mathcal{A}_x$ and $\mathcal{A}_y$ intersect, which contradicts the fact that $f$ is injective. The proof is complete.
The first proof also answers to the question that no mapping exists from $\mathbb{R}^n \rightarrow \mathbb{R}$.
Solution
In general there does not exist an $1-1$ and continuous mapping from $\mathbb{R}^n$ to $\mathbb{R}$ for $n \geq 2$. The first proof we are giving is an easy one. The second one is a bit more technical.
1st proof: Due to connectness we have that $f \left( \mathbb{R}^2 \right)= \mathcal{I}$ where $\mathcal{I}$ is an interval. Note that if we remove a point from the plane, it still remains connected. Having that in mind we observe that $f \left ( \mathbb{R}^2 \setminus f^{-1}\left ( a \right ) \right )$ is connected and equals $\mathcal{I} \setminus \left \{ a \right \}$ whereas this is not connected leading to a contradiction.
2nd proof: Define $\mathcal{A}_x = f(\{x\}\times\mathbb {R})$ for each $x \in\mathbb {R}$. Since $\{x\}\times\mathbb {R}$ is connected and $f$ is continuous, $\mathcal{A}_x$ must also be connected. Therefore it is an interval. It is not degenerate (a point), because $f$ is injective. Therefore each $\mathcal{A}_x$ contains an open interval $(\mathfrak{p}_x, \mathfrak{q}_x)$, where $\mathfrak{p}_x$ and $\mathfrak{q}_x$ are distinct rational numbers. Since there are only countably many such intervals $(\mathfrak{p}_x, \mathfrak{q}_x)$ and uncountably many sets $\mathcal{A}_x$, there exist $x \neq y$ such that $(\mathfrak{p}_x, \mathfrak{q}_x)=(\mathfrak{p}_y, \mathfrak{q}_y)$. Thus the sets $\mathcal{A}_x$ and $\mathcal{A}_y$ intersect, which contradicts the fact that $f$ is injective. The proof is complete.
The first proof also answers to the question that no mapping exists from $\mathbb{R}^n \rightarrow \mathbb{R}$.
Similarly, we can show that no such map from $\mathbb{S}^1$ to $\mathbb{R}$ exists.
ReplyDeleteIndeed,
If such a map $f$ existed, then $\mathbb{S}^1$ would be homeomorphic to its image, which is compact and connected in $\mathbb{R}^1$. But then this image would be a bounded closed interval, whose boundary would be nonempty while the boundary of $\mathbb{S}^1$ is empty, a contradiction.
I fell in love with the second proof!
ReplyDeleteGlad you liked it !! Take a tour of the site . You may find other posts interesting too !
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