In some sense, both perspectives are correct. There is the notion of density and the notion of a one-to-one correspondence.
First let us talk about the one-to-one correspondence. Given two sets $A$ and $B$, a function $f:A\rightarrow B$ is a one-to-one correspondence if $f$ satisfies the property that for every $b\in B$ there is a unique $a\in A$ such that $f(a)=b$.
As a concrete example given the sets $A=\{1,2,3\}$ and $B=\{a,b,c\}$ the function $f:A\rightarrow B$ where $f(1)=a$, $f(2)=b$, and $f(3)=c$ is a one-to-one correspondence.
An example of a function that is not a one-to-one correspondence would be something like $g:A\rightarrow B$ where $g(1)=a$, $g(2)=a$ and $g(3)=b$. This is not a one-to-one correspondence since there is no element of $A$ which maps to $c$ and we also have that the element mapping to $a$ is not unique (as both $1$ and $2$ map to $a$), so this function double fails in being a one-to-one correspondence.
Now we define two sets to have the same cardinality (size) if there is a one-to-one correspondence between them. That is given two sets $A$ and $B$ there is a function $f:A\rightarrow B$ that is a one-to-one correspondence. Thus, in our above example we have that $\vert \{1,2,3\}\vert=\{a,b,c\}$ since we had our one-to-one correspondence $f$.
There is a one-to-one correspondence between the even and whole numbers, we see this as follows. If we let $\mathbb{N}=\{1,2,3,...\}$ be the set of whole numbers, and we let $E=\{2,4,6,...\}$ be the set of even numbers, then we can construct the function $f:\mathbb{N}\rightarrow E$ by $f(n)=2n$, and this will give the one-to-one correspondence. We can check that it satisfies the two required properties, namely, we have that every even number $x\in E$ is of the form $x=2n$ (this is by definition of being even), and we need that given such an $x\in E$ if we write it as $2n$ where $n\in\mathbb{N}$, then this $n$ is unique. You can see this as if $n$ is not unique, then there is some $n'\neq n$ that is there say we could write $2n=2n'$, and rearranging this would imply that $2(n-n')=0$, so $n-n'=0$ and $n=n'$ which is a contradiction. Thus, we must have that the expression is unique, and we have our function is a one-to-one correspondence.
In this view, we have that $\vert \mathbb{N}\vert=\vert E\vert$.
Now to your point where it seems that there should be half as many even numbers as whole numbers. For this to be precise, we shall introduce the notion of density. The way I like to think about density is in terms of probability. That is if I asked for a random whole number what is the probability that it is even? Well, there are infinitely many options to choose from, so we can't just divide infinity by infinity. But what we can do instead is use calculus and take a limit, so let us define $[n]=\{1,2,3,...,n\}$. Then given some set $A\subset \mathbb{N}$, we define the density of $A$ as
$$
\lim_{n\rightarrow\infty}\frac{\vert A\cap[n]\vert}{n}
$$
If you think of this in terms of probability, it is asking the conditional probability question of given that the number I choose is less than of equal to $n$ what is the probability the number chosen is from the set $A$, and then we take the limit as $n$ goes to infinity to deal with the notion that $\mathbb{N}$ is infinite.
Thus, if we consider the density of $E$ inside $\mathbb{N}$, we will have that the probability sequence will be
$$0,\frac{1}{2},\frac{1}{3},\frac{1}{2},\frac{2}{5},\frac{1}{2},\frac{3}{7},\frac{1}{2},\frac{4}{9},...
$$
thus, we can see that
$$
\lim_{n\rightarrow\infty}\frac{\vert E\cap[n]\vert}{n}=\frac{1}{2}
$$
so we get that the density of the even numbers is $\frac{1}{2}$.
To conclude, there is a one-to-one correspondence between the whole numbers and even numbers; however, the density of the even numbers inside the set of whole numbers is $\frac{1}{2}$.
N, two items inEcould be matched (4n and 4n-2), so by your logic,Eis also twice as large asN. $\endgroup$