Infinite subsets of compact sets and limit points

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0x2a
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#1
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If E is an infinite subset of a compact set K, then E has a limit point in K

My proof goes like this:

Since E is an infinite set, a limit point exists. If x \in E, then x \in K so we are done. Let x be a limit point in E, such that  x \not \in E, then x \in E^c. If E is a proper subset of K (otherwise E = K, and since K is a compact set, it is closed and contains all its limit points are we are done), then x \in (K\setminus E )\cup K^c.

If x \in K^c then since K is a compact set and is closed, K^c is open. So x must be a interior point of K^c, which contradicts the fact that x is a limit point of E. So  x \in K\setminus E and x\in K.



So is this a sound proof?
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Smaug123
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(Original post by 0x2a)
If E is an infinite subset of a compact set K, then E has a limit point in K

My proof goes like this:

Since E is an infinite set, a limit point exists. If x \in E, then x \in K so we are done. Let x be a limit point in E, such that  x \not \in E, then x \in E^c. If E is a proper subset of K (otherwise E = K, and since K is a compact set, it is closed and contains all its limit points are we are done), then x \in (K\setminus E )\cup K^c.

If x \in K^c then since K is a compact set and is closed, K^c is open. So x must be a interior point of K^c, which contradicts the fact that x is a limit point of E. So  x \in K\setminus E and x\in K.



So is this a sound proof?
As I read this, I thought of the following things:

"Since E is an infinite set, a limit point exists." Why?
"K is compact implies K is closed." This is false: consider the cofinite topology on Z. All subsets of Z are compact, but only the finite ones (and Z) are closed. (It is true if K is also Hausdorff.)
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0x2a
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(Original post by Smaug123)
As I read this, I thought of the following things:

"Since E is an infinite set, a limit point exists." Why?
"K is compact implies K is closed." This is false: consider the cofinite topology on Z. All subsets of Z are compact, but only the finite ones (and Z) are closed. (It is true if K is also Hausdorff.)
Oh I forgot to mention that E \subset Y \subset X where X is a metric space. Does that make any difference? This is from an analysis book and not a topology one
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Smaug123
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(Original post by 0x2a)
Oh I forgot to mention that E \subset Y \subset X where X is a metric space. Does that make any difference? This is from an analysis book and not a topology one
That justifies "K is closed". I'm still not completely sure about "a limit point exists" - do you have a justification for that?
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0x2a
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(Original post by Smaug123)
That justifies "K is closed". I'm still not completely sure about "a limit point exists" - do you have a justification for that?
Well I'm stuck but if E has not limit points, then the set of limit points of E is the empty set, and so E is closed. Since E is a subset of a compact set and is closed, E is also a compact set.

Oh and if E has a limit point x, then x is a limit point of K, and since K is closed, K must contain all its limit points, so x \in K. Would this be a valid proof once I validate that a limit point of E exists at all?
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Smaug123
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(Original post by 0x2a)
Well I'm stuck but if E has not limit points, then the set of limit points of E is the empty set, and so E is closed. Since E is a subset of a compact set and is closed, E is also a compact set.

Oh and if E has a limit point x, then x is a limit point of K, and since K is closed, K must contain all its limit points, so x \in K. Would this be a valid proof once I validate that a limit point of E exists at all?
Actually, a limit point of any nonempty set exists, doesn't it? Take the sequence (x, x, …); this has limit x. That makes the original theorem completely trivial. I've presumably badly misunderstood something.
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0x2a
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(Original post by Smaug123)
Actually, a limit point of any nonempty set exists, doesn't it? Take the sequence (x, x, …); this has limit x. That makes the original theorem completely trivial. I've presumably badly misunderstood something.
It was defined that the neighbourhood of x must include a point not equal to x in all its neighbourhoods.
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