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Following was the homework question for my analysis class.

Given any sequence $x_n$ in a metric space $(X; d)$, and $x \in X$, consider the function $f : \mathbb N^\ast = \mathbb N \cup \{\infty\} \to X$ defined by $f(n) = x_n$, for all $n\in \mathbb N$, and $f(\infty) = x$.

Prove that there exists a metric on $\mathbb N^\ast$ such that the sequence $x_n$ converges to $x$ in $(X; d)$ if and only if the function $f$ is continuous.

Our teacher said that since $\infty$ is the only accumulation point of $\mathbb N^\ast$, if $f$ is continuous at $\infty$, it is continuous on $\mathbb N^\ast$. So, it is enough to show it is continuous at $\infty$.

But I did not understand why this is true. It is also possible that I misunderstood what my teacher meant. If you can tell me the statement is correct or not and why, I would be grateful.

Thanks in advance.

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1 Answer 1

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It's not clear to me what your teacher meant by $\infty$ being the only accumulation point of $\mathbb N^*$, since $\mathbb N^*$ is yet to be equipped with a metric and it makes no sense to speak of accumulation points before that.

A metric on $\mathbb N^*$ satisfying the requirement is induced by mapping $\mathbb N^*$ to $[0,1]$ using $n\mapsto1/n$, with $1/\infty:=0$, and using the canonical metric on $[0,1]$. Then all natural numbers are isolated points, and the open neighbourhoods of $\infty$ are the cofinite sets. The function $f$ is continuous if and only if the preimages of all open sets are open. The preimages of open sets not containing $x$ are open because they consist of a finite number of isolated points. The preimages of open sets containing $x$ are open because $x_n$ eventually ends up in every such open set, and thus the preimage is a cofinite set.

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Thank you for your answer. I still wonder the correctness of the statement. Assume that $(X,d)$ is a metric space and $a \in X$ is the only accumulation point. Then, can we say that if a function f : X→Y is continuos at a, then it is continuous on X? –  marvinthemartian Nov 17 '11 at 14:04
    
@marvin: Yes. At all other points $x\in X$, you can choose $\delta$ such that no other points are within $\delta$ of $x$, and clearly $|f(x)-f(x)|\lt\epsilon$, so that $\delta$ is good enough for any $\epsilon$. –  joriki Nov 17 '11 at 14:09

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