Proof is done

tomsch420 · May 31, 2024 · 0ecce90 · 0ecce90
1 parent 62acad7
commit 0ecce90
Showing 1 changed file with 19 additions and 11 deletions.
diff --git a/examples/product_spaces.ipynb b/examples/product_spaces.ipynb
@@ -681,8 +681,16 @@
     "    (A \\times B)^c &= (A^c \\times B) \\cup (A \\times B^c) \\cup (A^c \\times B^c) \\\\\n",
     "    &= (A^c \\times B) \\cup (A^c \\times B^c) \\cup (A \\times B^c) \\\\\n",
     "    &= ( A^c \\times (B \\cup B^c) ) \\cup   (A \\times B^c) \\\\\n",
-    "    &= (A^c \\times \\mathbb{B}) \\cup (A \\times B^C) \\square\n",
+    "    &= (A^c \\times \\mathbb{B}) \\cup (A \\times B^C) \\qquad \\square\n",
     "\\end{align*}\n",
+    "\n"
+   ],
+   "id": "511cdcad45f76bab"
+  },
+  {
+   "metadata": {},
+   "cell_type": "markdown",
+   "source": [
     "\n",
     "### Induction Step\n",
     "\n",
@@ -691,17 +699,17 @@
     "\\end{align*}\n",
     "Proof:\n",
     "\\begin{align*}\n",
-    "    (A \\times B \\times C)^c &= (A^c \\times B \\times C) \\cup (A \\times B^c \\times C) \\cup (A \\times B \\times C^c) \\cup \n",
-    "    (A^c \\times B^c \\times C) \\cup (A^c \\times B \\times C^c) \\cup (A \\times B^c \\times C^c) \\cup \n",
-    "    (A^c \\times B^c \\times C^c) \\\\\n",
-    "    &= (C \\times \\underbrace{(A^c \\times B) \\cup (A \\times B^c) \\cup (A^c \\times B^c))}_{\\text{Induction Assumption}} \\cup\n",
-    "    (C^c \\times  \\underbrace{(A^c \\times B) \\cup (A \\times B^c) \\cup (A^c \\times B^c))}_{\\text{Induction Assumption}} \\cup (A \\times B \\times C^c) \\\\\n",
-    "    &= (C \\times (A^c \\times \\mathbb{B}) \\cup (A \\times B^C)) \\cup \n",
-    "       (C^c \\times (A^c \\times \\mathbb{B}) \\cup (A \\times B^C)) \\cup (A \\times B \\times C^c)\\\\\n",
-    "    &= \n",
-    "\\end{align*}\n"
+    "    (A \\times B \\times C)^c &= (A \\times B^c \\times C) \\cup (A \\times B \\times C^c) \\cup (A \\times B^c \\times C^c)\\\\\n",
+    "     & \\cup \\, (A^c \\times B \\times C)\\cup(A^c \\times B^c \\times C) \\cup (A^c \\times B \\times C^c) \\cup (A^c \\times B^c \\times C^c) \\\\ \n",
+    "    &= (A \\times \\underbrace{((B^c \\times C) \\cup (B \\times C^c) \\cup (B^c \\times C^c))}_{\\text{Induction Assumption}}  \\\\\n",
+    "    &\\cup \\, (A^c \\times  \\underbrace{((B \\times C)\\cup(B^c \\times C) \\cup (B \\times C^c) \\cup (B^c \\times C^c))}_{\\mathbb{B} \\times \\mathbb{C}}) \\\\\n",
+    "    &= (A \\times (B^c \\times \\mathbb{C}) \\cup (B \\times C^C)) \\cup (A^c \\times \\mathbb{B} \\times \\mathbb{C}) \\\\\n",
+    "    &= (A^c \\times \\mathbb{B} \\times \\mathbb{C}) \\cup (A \\times B^C \\times \\mathbb{C} ) \\cup (A \\times B \\times C^c) \\qquad \\square\n",
+    "\\end{align*}\n",
+    "\n",
+    "Furthermore, take notice that the complement $(A \\times B \\times C)^c$ as expressed in this proof is a disjoint union."
    ],
-   "id": "511cdcad45f76bab"
+   "id": "4ca5ba06adf3eb6a"
   },
   {
    "cell_type": "markdown",