A sequence of rationals converging to an irrational point (proving that $mathbb Q$ is not locally compact)
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Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
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add a comment |
$begingroup$
Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
$endgroup$
add a comment |
$begingroup$
Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
$endgroup$
Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
25413
add a comment |
add a comment |
1 Answer
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Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
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1 Answer
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Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,240312
$endgroup$
add a comment |
$begingroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,240312
$endgroup$
add a comment |
$begingroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,240312
$endgroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,240312
answered Dec 9 '18 at 4:28
QuokaQuoka
1,240312
answered Dec 9 '18 at 4:28
QuokaQuoka
1,240312
answered Dec 9 '18 at 4:28
QuokaQuoka
1,240312
1,240312
add a comment |
add a comment |
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