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| -rw-r--r-- | Problem Set 2.page | 4 |
1 files changed, 2 insertions, 2 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index ed987ae..c0114b5 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -106,8 +106,8 @@ Proofs: -$2^{\mathbb{N}} = \mathbf{R}$ by writing a given real number $r$ as a (possibly infinite) set of natural numbers. For example, write $pi$ as the set of rational numbers $\{3,0.1,0.04,0.001,0.005,...\}$, then replace each number with the natural number it would map to under the bijection $\mathbb{Q}=\mathbf{N}$. -2. So we assume that there is some bijection f between all the elements in $X$ and $2^X$. This means that we can map every element in X to an element in $2^X$. For all the $x \in X atleast one x must map to a set in 2^X$ that does not contain itself. This is true since, we know that the empty set is a member of $2^X$, so one of the elements in X must map to the empty set. So lets look at all the $x \in X that map to sets in 2^X$ that do not contain the element x. All of these elements that map to an element in $2^X$ that doesn't contain themselves must be contained in another set that is a member of 2^X$. Since there is a bijection between X$ and 2^X$, there must be some other -$x \in X$ that points to this set we just created. However, if some other $x \in X$ points to this set, it also points to a set that does not contain itself. Therefore there must be some other set that contains it and all the other elements that point to sets that do not contain themselves. Each time you try to create this set, you create another element that does not point to itself. Therefore, in order to have a bijection between $X$ and $2^X$ all $x \. X$ must point to sets in $2^X$ that contain themselves, but we know that this isn't true since atleast one element in X should point to the empty set. +2. So we assume that there is some bijection f between all the elements in $X$ and $2^X$. This means that we can map every element in X to an element in $2^X$. For all the $x \in X$ atleast one x must map to a set in $2^X$ that does not contain itself. This is true since, we know that the empty set is a member of $2^X$, so one of the elements in X must map to the empty set.So lets look at all the $x \in X$ that map to sets in $2^X$ that do not contain the element x. All of these elements that map to an element in $2^X$ that do not contain themselves must be contained in another set that is a member of $2^X$. Since there is a bijection between $X$ and $2^X$, there must be some other +$x \in X$ that points to this set we just created. However, if some other $x \in X$ points to this set, it also points to a set that does not contain itself. Therefore there must be some other set that contains it. Each time you try to create this set, you create another element that does not point to itself. Therefore, in order to have a bijection between $X$ and $2^X$ all $x \. X$ must point to sets in $2^X$ that contain themselves, but we know that this isn't true since atleast one element in X should point to the empty set. # Comments |