Infinite subsets of compact sets and limit points

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?

(edited 9 years ago)

Reply 1

Smaug123

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.)

Reply 2

0x2a

Original post by Smaug123

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 :redface:

Reply 3

Smaug123

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 :redface:

That justifies "K is closed". I'm still not completely sure about "a limit point exists" - do you have a justification for that?

Reply 4

0x2a

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?

Reply 5

Smaug123

Original post by 0x2a

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.