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Fix typo

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Fix typos introduces more
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Zheyuan Wu
2024-12-03 11:20:59 -06:00
parent cbed1333ed
commit 9283c6b427
21 changed files with 213 additions and 44 deletions
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pages/Math4111/Math4111_L16.md
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@@ -146,7 +146,7 @@ This proves the claim.
By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\overline{E})\leq 2\epsilon+diam E$. So $diam(\overline{E})\leq diam E$.
(b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \emptyset$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0.
(b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \phi$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0.
EOP