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**Tinyboss** Rather than say there's an accumulation point in some neighborhood, you want to first show that (at least) one accumulation point exists in X (L in my post above), then say that any neighborhood of this point must contain infinitely many elements of Y, contrary to the hypothesis. You don't need any epsilonics here. (Or maybe you do...if this is all analysis and no topology...in that case, yeah, say that $\displaystyle (L-\varepsilon,L+\varepsilon)\cap X$ contains a point of Y distinct from L, for any epsilon, so in fact it contains infinitely many for any epsilon.)