fix typo

1 files changed, 1 insertions, 1 deletions

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 5ba091b..5d5cd25 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -107,7 +107,7 @@ 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 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 just this element and an $x \in X$ must point to this set. 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 $x \ in X$ 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.  
+$x \in X$ 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 just this element and an $x \in X$ must point to this set. 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 $x \in X$ must point to sets in $2^X$ that contain $x$, but we know that this isn't true since atleast one element in $X$ should point to the empty set.  
 
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