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Let $X$ be an arbitrary set. The non-empty open nsets are the complements offinite sets. We have to define also the empty set.

A sequence $x_n \to x$ converges as for the cofinite topology iff for each open neighbourhood $U$ of $x$, $U = X \setminus \{s\}$ for a $s$ with $x \neq s$, it holds that almost all $x_n$ are in $U$. So if $x \neq s$, thenalmost all $x_n \neq s$. If infinitelymany $x_n=s$ then $x=s$.

Have I understood that correctly?

So considering the case $x_n=\frac{1}{n}$:

Let $U$ be an open neighbourhood of $0$. Since $\Bbb R\setminus U$ must be finite, $U$ containsallbut a fininte numberof terms of the sequence. Therefore $x_n\rightarrow 0$.

Is that correct?

Mary Star
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1 Answers1

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Your characterization of convergence is not complete. A sequence $(x_n)$ is convergent in this topology iff no point $s$ is repeated infinitely many times in the sequence. In this case the sequence converges to every point of the space.

In particular, a sequence of distinct points like $(\frac 1 n )$ converges to every point.

Kavi Rama Murthy
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