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With the definition of continuity at a point as follows:

"Let $(X,d)$ and $(Y,\rho)$ be metric spaces. A function $f:X\rightarrow Y$ is continuous at the point $x \epsilon X$ if whenever ${x_n}$ is a sequence in X such that $x_n\rightarrow x$, then $f(x_n)\rightarrow f(x)$"

a theorem is stated in part:

"The function $f: X\rightarrow Y$ is continuous at $x \epsilon X$, if for $\epsilon\gt0$ there exists $\ \delta \gt 0$ such that whenever $z \epsilon X$ satisfies $d(x,z) \lt \delta$ , then $\rho(f(x),f(z)) \lt \epsilon$." [Note: I have purposely stated only the "if" part of the theorem, about which proof I have a question.] Now to the indirect proof and my question about it:

"Suppose that $f$ is not continous at $x$. Choose a sequence $\{x_k\}$ such that $\{x_k\}\rightarrow x$ but $f(x_k)$ does not converge to $f(x)$. Passing to a subsequence, we can assume that $\rho(f(x),f(x_k)) \ge \epsilon$ for some $\epsilon\gt 0$. Then there can exist no $\delta \gt0$, as in the condition of the theorem, so that the condition is violated."

Why do the authors (Gamelin and Greene "Introduction to Topology") pass to a subsequence ? This may seem minor point, but I am concerned that I am missing an essential.

  • $f(x_k) \to f(x)$ means precisely that for every $\epsilon > 0$, all but finitely many $x_k$ satisfy $\rho(f(x), f(x_k)) < \epsilon$. The negation of this statement is that for some $\epsilon > 0$, there are infinitely many $x_k$ such that $\rho(f(x), f(x_k)) \geq \epsilon$. "There are infinitely many" means precisely that there is a subsequence for which it's always true. –  Nov 04 '15 at 17:05
  • @Bungo Yes, indeed! –  Nov 04 '15 at 17:53

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Because otherwise the claim "we can assume that [for some $\epsilon>0$ and for all $k$,] $\rho(f(x), f(x_k))\ge\epsilon$" is false - maybe the sequence sometimes actually hits $f(x)$ repeatedly. For instance, maybe $f(x)=1$, and $f(x_k)=0$ if $k$ is even and $1$ if $k$ is odd. Then $\{f(x_k)\}$ does not converge to $f(x)$, but we frequently have $\rho(f(x), f(x_k))=0$.

(I edited your statement to make the fact that $\epsilon$ is independent of $k$ clearer, since this is an important point.)

Noah Schweber
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