# If f:X->Y is continuous and x is a limit point of subset A of X, is f(x) a limit point of f(A)?

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lapilofu (

4320)
February 27th, 2010

I heard a rumor that this is not true, but I can’t come up with a counterexample or a proof in favor of it.

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## 15 Answers

And here’s me, thinking i was the only one who heard that rumour…

Wolfram|Alpha isn’t sure how to compute an answer from your input

And neither is grumpyfish =)

Don’t believe everything you hear.

umm, yeah, sure, whatever you say, I completely understand

Obviously you have to diagulate the trimanuglas. Everyone knows that.

@lapilofu I’m sorry no one who understands topology has come along yet =) I did actually look up some of what you’re talking about, but it’s outside the math I’ve handled.

For those not familiar with your terminology (myself included)….*Can you rephrase your question in a more widely known and simpler way, please?*

I’d suggest employing the Socratic method to induce recollection of the truths you already know.

@lloydbird I’m afraid the question is really only for people who have some experience with topology or advanced set theory. To rephrase it in layman’s terms would require summarizing a course in set theory.

@lapilofu Oh, I see.

Then is there a website, that you could direct me to, where I (and whoever else wants to)

can access a *summary* of a “course in set theory”?

I’d like to be able to respond at some future date.

Rather than not at all.

@lloydbird Wikipidia’s pages on Set Theory and Topology are a pretty good sources, though there’s no replacement for a good textbook. Munkres’s “Toplogy: A First Course” is what I’ve been using and it’s pretty clear.

You seem offended by my response, and I’m sorry if that’s the case, I just honestly can’t think of how to simplify my question without typing pages of definitions and theorems.

Where is @finkelitis when you need him?

Let X be the Reals with the usual topology, and let Y be a space with the discrete topology. Let f:X->Y be constant, say f(x) = y for all x in X. Then f is continuous since the preimage of any subset B of Y under f is either X or empty depending on whether y is in B. Let A = [0,1); then 1 is a limit point of A, while f(1) is not a limit point of f(A) = {y}, since {y} is an open subset of f(A) that contains no member of f(A) distinct from y.

@lloydbird I am not in the least offended by your response. No apology needed.

Thanks for the tips. :-)

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