I am trying to solve a question from Pugh's Real Mathematical Analysis 2nd ed. The question seemed a little bit odd to me, and is as follows;
Construct a subset A ⊂ $\mathbb{R}$ and a continuous bijection $f: A → A$ that is not a homeomorphism.
Since the question does not imply anything about using different metrics, I suppose A is equipped with the usual metric of $\mathbb{R}$. In addition, I know that A must be noncompact. But I cannot go further to construct such a subset.
Best Answer
Let $A = \mathbb N \cup \{0\} \cup \{\frac{1}{n} \mid n \in \mathbb N \}$. Note that $0$ is the only non-isolated point of $A$.
Define $$f : A \to A, f(x) = \begin{cases} 0 & x= 0 \\ \frac{1}{2n} & x = \frac{1}{n}, n \in \mathbb N \\ m & x = 2m, m> 0 \\ \frac{1}{2m+1} & x = 2m+1,m > 0 \end{cases}$$ It is easy to verify that this map is a bijection. It is continuous in all $x \ne 0$ since these points are isolated. Let $\epsilon > 0$. If $\lvert x - 0 \rvert = x < \min(\epsilon,1)$, then either $x = 0$ or $x = \frac{1}{n}$ for some $n$. In both cases $f(x) = \frac{x}{2}$ and $\lvert f(x) - f(0) \rvert = \frac{x}{2} < \epsilon$. Therefore $f$ is continuous in $0$. If it were a homeomorphism, the closed subset $\mathbb N \subset A$ would be a closed image in $A$. But this is not true because $f(\mathbb N)$ has $0$ as cluster point.