Cardinality of set of actual continual features

The set of all $\mathbb{R\to R}$ continual features is $\mathfrak c$. Just how to show that? Exists any kind of bijection in between $\mathbb R^n$ and also the set of continual features?

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2019-05-04 17:40:58
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The cardinality goes to the very least that of the continuum due to the fact that every actual number represents a constant function. The cardinality goes to the majority of that of the continuum due to the fact that the set of actual continual features infuses right into the series room $R^{N}$ by mapping each continual function to its values on all the sensible factors. Given that the sensible factors are thick, this establishes the function.

The Schroeder-Bernstein theorem currently indicates the cardinality is specifically that of the continuum.

Keep in mind that after that the set of series of reals is additionally of the very same cardinality as the reals. This is due to the fact that if we have a series of binary depictions $.a_1a_2..., .b_1b_2..., .c_1c_2...$, we can splice them with each other using $.a_1 b_1 a_2 c_1 b_2 a_3...$ to make sure that a series of reals can be inscribed by one actual number.

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2019-05-08 04:58:07
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