T. Eskin
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 Mar 9 answered Classify each of the following sets as open, closed, or neither Mar 9 comment If, $\forall n\in\mathbb{N}$, $\lvert x_{n+1} - x_n \lvert\le a^n$ (for $a\in (0,1)$), then $x_n\to x$ @sequence. The first inequality is triangle inequality applied $m-n$ times and the equality after that is just expressing the partial sum from $n$ to $m-1$ as the difference of the same sum by noticing that it is the same as from $0$ to $m-1$ minus from $0$ to $n-1$. Then just use the geometric series formula. Mar 9 revised If, $\forall n\in\mathbb{N}$, $\lvert x_{n+1} - x_n \lvert\le a^n$ (for $a\in (0,1)$), then $x_n\to x$ added 316 characters in body Mar 9 answered If, $\forall n\in\mathbb{N}$, $\lvert x_{n+1} - x_n \lvert\le a^n$ (for $a\in (0,1)$), then $x_n\to x$ Mar 9 reviewed Reject Maximal ideals in rings of polynomials Mar 9 comment If $K$ is compact in $\mathbb{R}^n$ s. t. $B_r(\vec{x}) \cap K =\{\vec{x}\}$ then $K$ is finite @sequence. True, that is the definition. I don't know if I fully understand your question. By saying the word "cover" we already include the fact that $K$ is a subset of the union, by definition. So we don't necessarily have to write the union at any point during our argument if we use the term "cover". Mar 9 comment If $K$ is compact in $\mathbb{R}^n$ s. t. $B_r(\vec{x}) \cap K =\{\vec{x}\}$ then $K$ is finite @sequence. If $\{U_{i}\}_{i\in I}$ is any cover of $K$, then every subcover $\{U_{j}\}_{j\in J}$ automatically satisfies $$K\subseteq \bigcup_{j\in J}U_{j}\subseteq \bigcup_{i\in I}U_{i}.$$ The first inclusion comes from the word "cover" and the latter inclusion from the word "sub". The open cover here is $\{U_{i}\}_{i\in I}$ and not $\bigcup_{i\in I}U_{i}$. The latter is just a superset of $K$. In our problem, the finite subcover is special because each member of the cover covers only one point of $K$. This means that we can cover $K$ by covering only finitely many of its points. Mar 9 comment If $K$ is compact in $\mathbb{R}^n$ s. t. $B_r(\vec{x}) \cap K =\{\vec{x}\}$ then $K$ is finite @sequence. 1. There is no such thing as infinitesimally small radius. 2. Infimum is defined for collections of real numbers, and the collection I'm referring to is the infimum of the radii which are reals. 3. This helps because if $\{B_{r_{i}}(x_{i})\}_{i=1}^{n}$ is a cover of $K$ and $B_{r_{i}}(x_{i})\cap K=\{x_{i}\}$ for each $i=1,...,n$, then each ball covers only one point of $K$. There are finitely many balls, so this union covers only finitely many points of $K$. But it is a cover of $K$, so we have to conclude that $K$ has only finitely many points. Mar 9 answered Converging sequences with real numbers Mar 9 revised Converging sequences with real numbers added 27 characters in body Mar 9 reviewed Reject Explain why the following conjecture for $f(x)=[x]+(x-[x])^{[x]}$ is not correct? Mar 9 reviewed Approve Pullback of an invertible sheaf through an isomorphism Mar 9 revised If $K$ is compact in $\mathbb{R}^n$ s. t. $B_r(\vec{x}) \cap K =\{\vec{x}\}$ then $K$ is finite added 46 characters in body Mar 9 reviewed Approve functional analysis-density of spaces Mar 9 reviewed Reject Number of distinct dense subsets Mar 9 answered If $K$ is compact in $\mathbb{R}^n$ s. t. $B_r(\vec{x}) \cap K =\{\vec{x}\}$ then $K$ is finite Mar 9 reviewed Approve if the fibers of all points are finite then will the map be finite? Mar 9 answered Find a surjective function $f:\mathbb{N}\to \mathbb{Q}$ Mar 9 reviewed Reject Is it true that a 3rd order polynomial must have at least one real root? Mar 9 reviewed Approve False: if $C$ is closed then closure of interior of $C$ is equal to $C$?