- #1
mahler1
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Homework Statement .
Let ##S==\{\frac{1}{n}+\frac{1}{m}, n,m \in \mathbb N\}##, find all the accumulation points of ##S## and decide whether ##S## is compact or not. The attempt at a solution.
I've found that the points of the form ##\frac{1}{n}## and ##0## are accumulation points. For the first, given ##ε>0##, take the point ##\frac{1}{n}+\frac{1}{m}## such that ##m>\frac{1}{ε}##.
Then, ##|\frac{1}{n}-(\frac{1}{n}+\frac{1}{m})|=|\frac{1}{m})|<ε##. To prove ##0## is an accumulation point, given ##ε>0##, take ##n: n>\frac{1}{ε} \implies \frac{1}{n}<ε##.
How can I prove that these are the only accumulation points of ##S##?
For compactness, by the Heine Borel theorem, ##S## is compact ##\iff## ##S## is closed and bounded. But ##0 \in \overline S## and clearly ##0## doesn't exist in ##S##, this means ##S## is not closed so ##S## is not compact.
Let ##S==\{\frac{1}{n}+\frac{1}{m}, n,m \in \mathbb N\}##, find all the accumulation points of ##S## and decide whether ##S## is compact or not. The attempt at a solution.
I've found that the points of the form ##\frac{1}{n}## and ##0## are accumulation points. For the first, given ##ε>0##, take the point ##\frac{1}{n}+\frac{1}{m}## such that ##m>\frac{1}{ε}##.
Then, ##|\frac{1}{n}-(\frac{1}{n}+\frac{1}{m})|=|\frac{1}{m})|<ε##. To prove ##0## is an accumulation point, given ##ε>0##, take ##n: n>\frac{1}{ε} \implies \frac{1}{n}<ε##.
How can I prove that these are the only accumulation points of ##S##?
For compactness, by the Heine Borel theorem, ##S## is compact ##\iff## ##S## is closed and bounded. But ##0 \in \overline S## and clearly ##0## doesn't exist in ##S##, this means ##S## is not closed so ##S## is not compact.