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Let $X$ be a denumerable set and $Y$ an infinite subset of $X$.

There are two functions $g$, $h$ such that $g:X \sim \mathbb{N}$(since $X$ is denumerable), $h:Y\to \mathbb{N}$ and $h$ is defined by $h(y)$ = the number of elements in $\{1,2,3,···,g(y)\} \cap g(Y)$.

Then I have to prove that $h$ is bijection. How to prove it?

($g:X \sim \mathbb{N}$ means that $g:X \to \mathbb{N}$ is bijecion.)

Right_No
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2 Answers2

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Alternate proof.
As $X$ is denumberal there is a bijection $f:X \to \mathbb{N}$.
As $Y$ is infinite there is an injection $g:\mathbb{N} \to Y$.
Thus $|\mathbb{N}| = |g(\mathbb{N})| \le |Y| \le |X| = |\mathbb{N}|$.
So $|Y| = |\mathbb{N}|$, $Y$ is denumerable.

Aryaman Maithani
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William Elliot
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First, we show that $h$ is an injection. This is easy to do.
Let $y_1, y_2 \in Y$ with $y_1 \neq y_2$. Then, we have $g(y_1) \neq g(y_2)$. (Since $g$ is a bijection and in particular, an injection.)

WLOG, assume that the $g(y_1) < g(y_2)$. In this case, we have $\{1, \ldots, g(y_1)\} \subsetneq \{1, \ldots, g(y_2)\}.$ Moreover, $g(y_2) \in g(Y)$ and thus, $\{1, \ldots, g(y_1)\} \cap g(Y) \subsetneq \{1, \ldots, g(y_2)\} \cap g(Y).$
(Simply put: $g(y_2)$ is an element of $\{1, \ldots, g(y_2)\} \cap g(Y)$ but not $\{1, \ldots, g(y_1)\} \cap g(Y)$.)

This tells us that $h(y_1) < h(y_2)$ which proves injectivity of $h$.

Before showing the surjectivity of $h$, we note the following lemma:

Lemma. Given any $n \in \mathbb{N}$, $g(Y)$ has at least $n$ elements.

The proof should be fairly straightforward.

Now, we show the surjectivity of $h$. Let $n \in \mathbb{N}.$
Note that $g(Y)$ is a subset of $\mathbb{N}$ and hence, is well-ordered.
Let $k_n$ be the $n^{\text{th}}$ smallest element of $g(Y)$. Then, $k_n = g(y_n)$ for some (unique) $y_n \in Y$.

Then, verify that $h(y_n) = n.$ This conlcudes the proof.

Aryaman Maithani
  • 15,771