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prove that $(M,d)$ is compact $leftrightarrow$ every infinite subset of $M$ has a limit point.

Help understanding why a complete, totally bounded metric space implies every infinite subset has a limit pointSome equivalent formulations of compactness of a metric spaceTotally bounded, complete $implies$ compactIf a metric space has the limit point property, is it separable? (ZF + AC$_omega$)If $X$ is complete, then every subset of $X$ that is closed and totally bounded is also compact. Does the converse hold?Subsequences of a sequence converging and the Bolzano Weierstrass theoremA metric space is complete when every closed and bounded subset of it is compactProve that a metric space is sequentially compact iff every infinite subset has a cluster pointProve a metric space in which every infinite subset has a limit point is compact.(Question about one particular proof)Metric space with every infinite set having a limit point

0

1

$begingroup$

Here's my attempt
$rightarrow$:

$textbfLet A be a infinite subset of M$.
$M$ is compact implies that $M$ is totally bounded and complete.
Since $A subset M$, $A$ is also totally bounded. And by the Bolzano-Weierstrass Theorem (Every infinite, totally bounded subset of a complete metric space $M$ has a limit point in $M$) $A$ has a limit point in $M$.

$leftarrow$:

Let $(x_n)$ be a sequence in $M$ and $A = x_n: ngeq1$ be its Range. I want to show that $(x_n)$ has a convergent subsequence. Since $A$ is an infinite subset of $M$ it has a limit point $x$ in $M$. That is $exists$ $(y_n) subset A$ such that $(y_n)rightarrow x$.
Since $(y_n)subset A$ then $(y_n)$ must be a subsequence of $(x_n)$. That is $(x_n)$ has a convergent subsequence in $M$, which implies that $M$ is compact.

What do you think?...Thanks!

proof-verification metric-spaces

share|cite|improve this question

asked 2 days ago

tnt235711tnt235711

1469

$endgroup$

add a comment | 

0

1

$begingroup$

Here's my attempt
$rightarrow$:

$textbfLet A be a infinite subset of M$.
$M$ is compact implies that $M$ is totally bounded and complete.
Since $A subset M$, $A$ is also totally bounded. And by the Bolzano-Weierstrass Theorem (Every infinite, totally bounded subset of a complete metric space $M$ has a limit point in $M$) $A$ has a limit point in $M$.

$leftarrow$:

Let $(x_n)$ be a sequence in $M$ and $A = x_n: ngeq1$ be its Range. I want to show that $(x_n)$ has a convergent subsequence. Since $A$ is an infinite subset of $M$ it has a limit point $x$ in $M$. That is $exists$ $(y_n) subset A$ such that $(y_n)rightarrow x$.
Since $(y_n)subset A$ then $(y_n)$ must be a subsequence of $(x_n)$. That is $(x_n)$ has a convergent subsequence in $M$, which implies that $M$ is compact.

What do you think?...Thanks!

proof-verification metric-spaces

share|cite|improve this question

asked 2 days ago

tnt235711tnt235711

1469

$endgroup$

add a comment | 

$begingroup$

Here's my attempt
$rightarrow$:

$textbfLet A be a infinite subset of M$.
$M$ is compact implies that $M$ is totally bounded and complete.
Since $A subset M$, $A$ is also totally bounded. And by the Bolzano-Weierstrass Theorem (Every infinite, totally bounded subset of a complete metric space $M$ has a limit point in $M$) $A$ has a limit point in $M$.

$leftarrow$:

Let $(x_n)$ be a sequence in $M$ and $A = x_n: ngeq1$ be its Range. I want to show that $(x_n)$ has a convergent subsequence. Since $A$ is an infinite subset of $M$ it has a limit point $x$ in $M$. That is $exists$ $(y_n) subset A$ such that $(y_n)rightarrow x$.
Since $(y_n)subset A$ then $(y_n)$ must be a subsequence of $(x_n)$. That is $(x_n)$ has a convergent subsequence in $M$, which implies that $M$ is compact.

What do you think?...Thanks!

proof-verification metric-spaces

share|cite|improve this question

asked 2 days ago

tnt235711tnt235711

1469

$endgroup$

share|cite|improve this question

asked 2 days ago

tnt235711tnt235711

1469

share|cite|improve this question

asked 2 days ago

tnt235711tnt235711

1469

asked 2 days ago

tnt235711tnt235711

1469

asked 2 days ago

tnt235711tnt235711

1469