If $X\subset\mathbb{R}$ and $Y\subset\mathbb{R}^2$ are homeomorphic, then the interior of $Y$ is empty 
Let $X\subset\mathbb{R}$ and $Y\subset\mathbb{R}^2$. If $X$ and $Y$ are homeomorphic, then $int(Y)=\emptyset$.

My approach: If $X$ and $Y$ are homeomorphic, then exist a function $f$, such that $f:X\to Y$ is continuous with inverse also continuous. Then $f:X\to f(X)\subset \mathbb{R}^{2}$, and $f^{-1}:Y\to f(Y)\subset\mathbb{R}$. This mean that $f(Y)$ is a interval. Why $int(Y)$ is empty?? Regards.
 A: Suppose there exists $x\in\text{int}\, Y$. Since $\text{int}\, Y$ is an open and nonempty subset of $\mathbb R^2$, it has an uncountable number of points. Let $B$ be a connected component of $\text{int}\,Y$ and let $A=B\setminus\{x\}$. We know that $A$ is nonempty. I'll leave it up to you to show that $A$ is also connected. Let $f:Y\to X$ be a homeomorphism. Consider the restriction $g=f|_{\text{int} A}$. Then $g(A)=f(B)\setminus\{f(x)\}$. Since $f$ and $g$ are homeomorphisms, these are both open sets. Since $A$ is connected, so is $g(A)$. Since $B$ is connected, so is $f(B)$, so it is an open interval. But $f(B)\setminus\{x\}$ is missing a point, so it is not connected, thus these two sets cannot be equal. This is a contradiction, so $\text{int}\,Y=\emptyset$.
A: Suppose that $\operatorname{int} Y\neq\varnothing$. Then, $Y$ contains a circle. In fact, $Y$ contains many circles: there exists some $\overline{\varepsilon}>0$ and $(a_0,b_0)\in\operatorname{int}Y$ such that if $0<\varepsilon<\overline{\varepsilon}$, then $$\{(a,b)\in\mathbb R^2\,|\,(a-a_0)^2+(b-b_0)^2\leq\varepsilon^2\}\subseteq Y.$$ For any such $\varepsilon$, take the boundary of the corresponding circle: $$C_{\varepsilon}\equiv\{(a,b)\in\mathbb R^2\,|\,(a-a_0)^2+(b-b_0)^2=\varepsilon^2\}$$ and consider $f^{-1}(C_{\varepsilon})$. Since $C_{\varepsilon}$ is connected in $\mathbb R^2$ and $f^{-1}$ is continuous, it follows that $f^{-1}(C_{\varepsilon})$ is also connected in $\mathbb R$. Therefore, $f^{-1}(C_{\varepsilon})$ is a (non-empty and non-degenerate) interval in $\mathbb R$. Now, $C_{\varepsilon'}$ and $C_{\varepsilon''}$ are disjoint if $\varepsilon'\neq\varepsilon''$ (they are boundaries of circles of different radii with the same center) and thus so are $f^{-1}(C_{\varepsilon'})$ and $f^{-1}(C_{\varepsilon''})$, given that $f^{-1}$ is injective. This implies that the collection $$\{f^{-1}(C_{\varepsilon})\}_{0<\varepsilon<\overline{\varepsilon}}$$ is an uncountable family of disjoint non-degenerate intervals of $\mathbb R$, which is impossible.
