Is $\aleph_0 = \mathfrak{c}$?

As we all know that $2^{\aleph_0}=\mathfrak{c}$, where $\aleph_0$ (pronounced as aleph-not) is the cardinality of the set of natural number(countably infinite) and '$\mathfrak{c}$' is the cardinality of the set of real numbers,which is uncountably infinite. From the above given formula one can easily conclude that $\mathfrak{c}$ can never be equal to $\aleph_0$.

However,when we consider infinite sets ,there is a very well known theorem that an infinite set is numerically equivalent to a subset of itself. Considering the set of natural numbers which is a subset of real numbers, then the above theorem states that the cardinality of the set of natural numbers is equal to the cardinality of real numbers. Therefore we can conclude that $\aleph_0$ = $\mathfrak{c}$. Is it that the definition of subset changes when we consider an uncountably infinite set? When we consider subsets of an uncountably infinite set,does it implies that the subsets are also uncountably infinite?

• An infinite set is equicardinal with some proper subset of it, but not every subset of it – eepperly16 Dec 6 '17 at 5:21
• I don't understand the downvotes. This seems like a perfectly useful and clear question. – Austin Mohr Dec 6 '17 at 5:22
• The power of the continuum is written $\mathfrak c$, which you get with \mathfrak c – Ross Millikan Dec 6 '17 at 5:23
• Thanks @RossMillikan for the information. – harshpsyche Dec 6 '17 at 5:25
• @AustinMohr My guess is some dismissed it as set theory denial crackpottery that occasionally pops up here cause of language like "christened by Cantor". I had that thought at first, but reading closely, it's far from clear this is the case (I didn't downvote). Or could be what eepperly just said. – spaceisdarkgreen Dec 6 '17 at 5:27

If it did, you'd have to conclude that $\{1,2,3\}$ is an infinite set, as it is a proper subset of the infinite set $\mathbb N$.