A continuous bijection can't exists because $[0,1]$ is a compact set and continuous functions send compacts in compacts. You can look for a non-continuous bijection, that exists because $[0,1]$ and $\mathbb R$ have the same cardinality.

I am having problems being able to formally demonstrate when a function is bijective (and therefore, surjective and injective).

So if we can find a nice bijection between the real numbers the infinite sequences of natural numbers we are about done. Now, we know that $\mathbb{N^N}$ can be identified with the real numbers, in fact continued fractions form a bijection between the irrationals and $\mathbb{N^N}$.

2015年11月9日 · $\begingroup$ I'm studying for an exam, so if this question is asked, I would first prove there exists bijections from $\mathbb{Q} \to \mathbb{N}$, $\mathbb{N} \to \mathbb{N} \times \mathbb{N}$, and $\mathbb{N} \times \mathbb{N} \to \mathbb{Q} \times \mathbb{Q}$ and then compose those functions and use the proof that the composition of bijections is bijective?

2020年4月15日 · A bijection is an isomorphism in the category of Sets. When the word "isomorphism" is used, it is always referred to the category you are working in. I will list some categories including their typical names for isomorphism: Sets: Bijection; Groups: Isomorphism; Top: Homeomorphism; Differentiable Manifolds: Diffeomorphism; Riemannian Manifolds ...

2010年10月24日 · Step Two: We showed there exists a bijection between $\mathbb{N}$ and $\mathbb{Q}^{+}$. We now attempt to show there exists an explicit bijection between $\mathbb{N}$ and $\mathbb{Q}$. Using the work done in Step One, it appears easier to first create a bijection between $\mathbb{Z}$ and $\mathbb{Q}$. The reason for doing so is …

2014年3月2日 · It is well-known that, by Banach theorem, every continuous, linear and bijective operator between Banach spaces is a isomorphism. There must be a linear bijective and discontinuous operator between

2015年4月1日 · A bijection f with domain X (indicated by f: X → Y in functional notation) also defines a relation starting in Y and going to X. Properties (3) and (4) of a bijection say that this inverse relation is a function with domain Y.

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2014年12月12日 · For example, if the sets are groups, then an isomorphism is a bijection that preserves the operation in the groups: $\varphi(ab) = \varphi(a)\varphi(b)$. As another example, if the sets are vector spaces, then an isomorphism is a bijection that preserves vector addition and scalar multiplication.