Learn the Interior of a set of complete Theorems with 6 Proof

 \displaystyle \begin{array}{l}\mathbf{Let}\text{ }(\mathbf{X},\tau )\text{ }\mathbf{be}\text{ }\mathbf{a}\text{ }\mathbf{topological}\text{ }\mathbf{space}\text{ }\mathbf{and}\text{ }\mathbf{A}\subseteq \mathbf{X}\text{ }\mathbf{then},\\(1)\text{ }{{\phi }^{o}}=\phi \text{          (2) }{{\text{X}}^{o}}=\text{X}\\(3)\text{ A is open iff  }{{\text{A}}^{o}}=\text{A}\\\text{(4) (}{{\text{A}}^{o}}{{)}^{o}}={{\text{A}}^{o}}\\(5)\text{  }{{\text{A}}^{o}}\subseteq \text{A}\\\text{(6) }{{\text{A}}^{o}}\text{ is the largest open subset of A}\\\mathbf{Proof}:1\\\text{since }\phi \text{ is open }\\\phi \text{ is an open subset of }\phi \\but\\{{\phi }^{o}}\text{ is the largest open subset of }\phi \text{ }\\\Rightarrow \phi \subseteq \text{ }{{\phi }^{o}}\subseteq \phi \\\Rightarrow \phi \subseteq \text{ }{{\phi }^{o}}\text{ and }{{\phi }^{o}}\subseteq \phi \\\Rightarrow {{\phi }^{o}}=\phi \\\mathbf{Proof}:2\\\text{since X is open }\\\text{X is an open subset of X}\\but\\{{X}^{o}}\text{ is the largest open subset of X }\\\Rightarrow X\subseteq \text{ }{{\text{X}}^{o}}\subseteq X\\\Rightarrow X\subseteq \text{ }{{\text{X}}^{o}}\text{ and }{{\text{X}}^{o}}\subseteq X\\\Rightarrow {{X}^{o}}=X\\\mathbf{Proof}:3\\\text{suppose }{{\text{A}}^{o}}\text{=A Then, we have to show that  A is open }\\{{\text{A}}^{o}}\text{=union of all open subsets of A}\\\text{So, }{{\text{A}}^{o}}\text{ is open }\\\Rightarrow \text{A is open }\\\mathbf{Conversely}\\\text{suppose A is open Then, we have to show that }{{\text{A}}^{o}}\text{=A}\\\because \text{A}\subseteq \text{A}\\\text{A is open subset of A}\\\text{but }{{\text{A}}^{o}}\text{ is the largest open subset of A}\\\Rightarrow \text{A}\subseteq \text{ }{{\text{A}}^{o}}\subseteq \text{A}\\\Rightarrow \text{A}\subseteq \text{ }{{\text{A}}^{o}}\text{ and }{{\text{A}}^{o}}\subseteq \text{A}\\\Rightarrow {{\text{A}}^{o}}=\text{A}\\\mathbf{Proof}:4\\\text{As we know that }{{\text{A}}^{o}}=\text{A iff A is open}\\\text{Since }{{\text{A}}^{o}}\text{ is open }\\\text{Int(}{{\text{A}}^{o}}\text{)=Int(A)}\\{{\text{(}{{\text{A}}^{o}})}^{o}}={{\text{A}}^{o}}\\\mathbf{Proof}:5\\{{\text{A}}^{o}}\text{=union of all open subsets of A}\\{{\text{A}}^{o}}\text{=}\bigcup\limits_{{i=1,2,3,...}}{{{{\text{H}}_{i}}}}\text{ where }{{\text{H}}_{i}}\in \tau ,{{\text{H}}_{i}}\subseteq \text{A}\\\text{As }{{\text{H}}_{i}}\subseteq \text{A}\\\Rightarrow \bigcup\limits_{{i=1,2,3,...}}{{{{\text{H}}_{i}}}}\subseteq \text{A}\\{{\text{A}}^{o}}\text{=A                 }\because {{\text{A}}^{o}}\text{=}\bigcup\limits_{{i=1,2,3,...}}{{{{\text{H}}_{i}}}}\\\mathbf{Proof}:6\\{{\text{A}}^{o}}\text{=union of all open subsets of A}\\\text{As union of open sets is open}\\{{\text{A}}^{o}}\text{ is open subset of A}\\\text{Suppose H is any open subset of A}\\\Rightarrow \text{H}\subseteq \text{union of all open subsets of A}\\\Rightarrow \text{H}\subseteq {{\text{A}}^{o}}\\\Rightarrow \text{ }{{\text{A}}^{o}}\text{ is the largest open subset of A}\end{array}