Mixing
Why is my proof that a space is closed incorrect?
A metric space $X$ is closed if every convergent sequence $(x_n)$ in $X$ converges in the space. A space is closed if it contains all its boundary points, so then I thought one could show a space is closed by showing that an arbitrary point in the space is a boundary point. This could be done by taking an arbitrary point $x in X$, and showing that for every $epsilon > 0 $ one can find a point $ynotin X $, such that $d(x, y) < epsilon$.
I understood this is an invalid way to show a space is closed, since with this method I managed to show that $c_{00}$ (the space of sequences thats eventually only zeroes) is closed. But $c_{00}$ is not closed, so my proof method above must clearly be wrong.
Where is my proof method wrong?
metric-spaces
asked Nov 22 '18 at 11:59
landigiolandigio
143
add a comment |
A metric space $X$ is closed if every convergent sequence $(x_n)$ in $X$ converges in the space. A space is closed if it contains all its boundary points, so then I thought one could show a space is closed by showing that an arbitrary point in the space is a boundary point. This could be done by taking an arbitrary point $x in X$, and showing that for every $epsilon > 0 $ one can find a point $ynotin X $, such that $d(x, y) < epsilon$.
I understood this is an invalid way to show a space is closed, since with this method I managed to show that $c_{00}$ (the space of sequences thats eventually only zeroes) is closed. But $c_{00}$ is not closed, so my proof method above must clearly be wrong.
Where is my proof method wrong?
metric-spaces
asked Nov 22 '18 at 11:59
landigiolandigio
143
Well you'd have to show us the proof in order to see what's wrong. As a simpler counterexample consider $x=1$ and $X=[0,2]subseteq mathbb{R}$ with the induced metric from $mathbb{R}$.
– Kolja
Nov 22 '18 at 12:03
1
The fact that every point is a boundary point does not imply that a set is closed. Consider the unit circle (in the plane) with a single point removed. All the remaining points are boundary points. So is the removed point. Therefore the set does not contain all its boundary points. Hence it is not closed.
– Jyrki Lahtonen
Nov 22 '18 at 12:04
2
Also, note that you should be proving that $X$ contains all it's boundary points, not that all the points of $X$ are boundary points. Try to see the difference between these two.
– Kolja
Nov 22 '18 at 12:06
That is exactly what the OP said in the last phrase of the first paragraph. @JyrkiLahtonen
– nicomezi
Nov 22 '18 at 12:09
2
Every metric space is closed (in itself) . What do you mean by saying that $c_{00}$ is not closed. Are you talking about completeness of metric spaces? F or me the entire question makes little sense.
– Kavi Rama Murthy
Nov 22 '18 at 12:16
add a comment |
A metric space $X$ is closed if every convergent sequence $(x_n)$ in $X$ converges in the space. A space is closed if it contains all its boundary points, so then I thought one could show a space is closed by showing that an arbitrary point in the space is a boundary point. This could be done by taking an arbitrary point $x in X$, and showing that for every $epsilon > 0 $ one can find a point $ynotin X $, such that $d(x, y) < epsilon$.
I understood this is an invalid way to show a space is closed, since with this method I managed to show that $c_{00}$ (the space of sequences thats eventually only zeroes) is closed. But $c_{00}$ is not closed, so my proof method above must clearly be wrong.
Where is my proof method wrong?
metric-spaces
asked Nov 22 '18 at 11:59
landigiolandigio
143
A metric space $X$ is closed if every convergent sequence $(x_n)$ in $X$ converges in the space. A space is closed if it contains all its boundary points, so then I thought one could show a space is closed by showing that an arbitrary point in the space is a boundary point. This could be done by taking an arbitrary point $x in X$, and showing that for every $epsilon > 0 $ one can find a point $ynotin X $, such that $d(x, y) < epsilon$.
I understood this is an invalid way to show a space is closed, since with this method I managed to show that $c_{00}$ (the space of sequences thats eventually only zeroes) is closed. But $c_{00}$ is not closed, so my proof method above must clearly be wrong.
Where is my proof method wrong?
metric-spaces
metric-spaces
asked Nov 22 '18 at 11:59
landigiolandigio
143
asked Nov 22 '18 at 11:59
landigiolandigio
143
asked Nov 22 '18 at 11:59
landigiolandigio
143
asked Nov 22 '18 at 11:59
landigiolandigio
143
asked Nov 22 '18 at 11:59
landigiolandigio
143
143
Well you'd have to show us the proof in order to see what's wrong. As a simpler counterexample consider $x=1$ and $X=[0,2]subseteq mathbb{R}$ with the induced metric from $mathbb{R}$.
– Kolja
Nov 22 '18 at 12:03
1
The fact that every point is a boundary point does not imply that a set is closed. Consider the unit circle (in the plane) with a single point removed. All the remaining points are boundary points. So is the removed point. Therefore the set does not contain all its boundary points. Hence it is not closed.
– Jyrki Lahtonen
Nov 22 '18 at 12:04
2
Also, note that you should be proving that $X$ contains all it's boundary points, not that all the points of $X$ are boundary points. Try to see the difference between these two.
– Kolja
Nov 22 '18 at 12:06
That is exactly what the OP said in the last phrase of the first paragraph. @JyrkiLahtonen
– nicomezi
Nov 22 '18 at 12:09
2
Every metric space is closed (in itself) . What do you mean by saying that $c_{00}$ is not closed. Are you talking about completeness of metric spaces? F or me the entire question makes little sense.
– Kavi Rama Murthy
Nov 22 '18 at 12:16
add a comment |
Well you'd have to show us the proof in order to see what's wrong. As a simpler counterexample consider $x=1$ and $X=[0,2]subseteq mathbb{R}$ with the induced metric from $mathbb{R}$.
– Kolja
Nov 22 '18 at 12:03
1
The fact that every point is a boundary point does not imply that a set is closed. Consider the unit circle (in the plane) with a single point removed. All the remaining points are boundary points. So is the removed point. Therefore the set does not contain all its boundary points. Hence it is not closed.
– Jyrki Lahtonen
Nov 22 '18 at 12:04
2
Also, note that you should be proving that $X$ contains all it's boundary points, not that all the points of $X$ are boundary points. Try to see the difference between these two.
– Kolja
Nov 22 '18 at 12:06
That is exactly what the OP said in the last phrase of the first paragraph. @JyrkiLahtonen
– nicomezi
Nov 22 '18 at 12:09
2
Every metric space is closed (in itself) . What do you mean by saying that $c_{00}$ is not closed. Are you talking about completeness of metric spaces? F or me the entire question makes little sense.
– Kavi Rama Murthy
Nov 22 '18 at 12:16
Well you'd have to show us the proof in order to see what's wrong. As a simpler counterexample consider $x=1$ and $X=[0,2]subseteq mathbb{R}$ with the induced metric from $mathbb{R}$.
– Kolja
Nov 22 '18 at 12:03
Well you'd have to show us the proof in order to see what's wrong. As a simpler counterexample consider $x=1$ and $X=[0,2]subseteq mathbb{R}$ with the induced metric from $mathbb{R}$.
– Kolja
Nov 22 '18 at 12:03
1
1
The fact that every point is a boundary point does not imply that a set is closed. Consider the unit circle (in the plane) with a single point removed. All the remaining points are boundary points. So is the removed point. Therefore the set does not contain all its boundary points. Hence it is not closed.
– Jyrki Lahtonen
Nov 22 '18 at 12:04
The fact that every point is a boundary point does not imply that a set is closed. Consider the unit circle (in the plane) with a single point removed. All the remaining points are boundary points. So is the removed point. Therefore the set does not contain all its boundary points. Hence it is not closed.
– Jyrki Lahtonen
Nov 22 '18 at 12:04
2
2
Also, note that you should be proving that $X$ contains all it's boundary points, not that all the points of $X$ are boundary points. Try to see the difference between these two.
– Kolja
Nov 22 '18 at 12:06
Also, note that you should be proving that $X$ contains all it's boundary points, not that all the points of $X$ are boundary points. Try to see the difference between these two.
– Kolja
Nov 22 '18 at 12:06
That is exactly what the OP said in the last phrase of the first paragraph. @JyrkiLahtonen
– nicomezi
Nov 22 '18 at 12:09
That is exactly what the OP said in the last phrase of the first paragraph. @JyrkiLahtonen
– nicomezi
Nov 22 '18 at 12:09
2
2
Every metric space is closed (in itself) . What do you mean by saying that $c_{00}$ is not closed. Are you talking about completeness of metric spaces? F or me the entire question makes little sense.
– Kavi Rama Murthy
Nov 22 '18 at 12:16
Every metric space is closed (in itself) . What do you mean by saying that $c_{00}$ is not closed. Are you talking about completeness of metric spaces? F or me the entire question makes little sense.
– Kavi Rama Murthy
Nov 22 '18 at 12:16
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Well you'd have to show us the proof in order to see what's wrong. As a simpler counterexample consider $x=1$ and $X=[0,2]subseteq mathbb{R}$ with the induced metric from $mathbb{R}$.
– Kolja
Nov 22 '18 at 12:03
1
The fact that every point is a boundary point does not imply that a set is closed. Consider the unit circle (in the plane) with a single point removed. All the remaining points are boundary points. So is the removed point. Therefore the set does not contain all its boundary points. Hence it is not closed.
– Jyrki Lahtonen
Nov 22 '18 at 12:04
2
Also, note that you should be proving that $X$ contains all it's boundary points, not that all the points of $X$ are boundary points. Try to see the difference between these two.
– Kolja
Nov 22 '18 at 12:06
That is exactly what the OP said in the last phrase of the first paragraph. @JyrkiLahtonen
– nicomezi
Nov 22 '18 at 12:09
2
Every metric space is closed (in itself) . What do you mean by saying that $c_{00}$ is not closed. Are you talking about completeness of metric spaces? F or me the entire question makes little sense.
– Kavi Rama Murthy
Nov 22 '18 at 12:16