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Suppose that $f:X \rightarrow Y$ is continuous. Suppose that we have some sequence $\{x_n \} \in X$ such that $f(x_n) \rightarrow f(x)$. Assume that $X$ is first countable (at every point $x \in X$, there is a countable basis set $B$ such that every open neighbour of $x$ contains at least one of the elements in $B$). Show that $x_n \rightarrow x$.

In Theorem 21.3 of Munkres, he showed that given any set $A \subset X$, once $f(\bar{A}) \subset \overline{f(A)}$ is shown, we are done. I don't see how showing that $f(\bar{A}) \subset \overline{f(A)}$ implies the result in the previous paragraph.

As shown by the comments, the claims of the previous paragraph are false.

I feel that I am misunderstanding the theorem in Munkres. I will just cite the theorem here:

(Theorem 21.3) Let $f : X \rightarrow Y$. If the function $f$ is continuous, then for every convergent sequence $x_n \rightarrow x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. The converse holds if $X$ is metrizable.

Isn't the converge to theorem 21.3 the statement: Let $f : X \rightarrow Y$. If the function $f$ is continuous, then for every sequence $x_n$ in $X$ such that $f(x_n)$ converges to $f(x)$, the sequence $x_n \rightarrow x$ in $X$?

darkgbm
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    Err, $\sin \frac{1}{n} \to \sin \pi$. – Daniel Fischer Nov 11 '16 at 22:41
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    This is false. Consider $f(x) = x^2$ and $x_n = (-1)^n$. Then $f(x_n) \to 1$ but clearly $x_n$ does not have a limit. More generally, if $f$ is any continuous but not injective function, similar counterexamples can be made. – davidlowryduda Nov 11 '16 at 22:42
  • @DanielFischer, feel that I am not understanding the statement correctly. I have cited the original statement in Munkres. – darkgbm Nov 12 '16 at 19:07
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    I see. It's unclear what "the converse" is, unless one knows what it should be. The converse that Munkres is talking about is "Let $f\colon X \to Y$. (Where $X$ is a metrisable space [that's more than is required for the conclusion], and $Y$ a topological space.) If for every sequence $(x_n)$ in $X$ with $x_n \to x$ we have $f(x_n) \to f(x)$, then $f$ is continuous." The thing is that sequential continuity [that is the property $(x_n \to x) \implies (f(x_n) \to f(x))$ for all sequences] is generally a weaker property than continuity, but for metrisable spaces, they are equivalent. – Daniel Fischer Nov 12 '16 at 19:13
  • @DanielFischer How does showing $f(\bar{A}) \subset \overline{f(A)}$ implies that $f$ is continuous? – darkgbm Nov 12 '16 at 20:04
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    Let $B \subset Y$ be closed. Define $A = f^{-1}(B)$. Then $f(A) \subset B$, and by what one has shown it follows that $f(\overline{A}) \subset \overline{f(A)} \subset \overline{B} = B$, whence $\overline{A}\subset f^{-1}(B) = A$, i.e. $A$ is closed. Thus $(\forall A)\bigl(f(\overline{A}) \subset \overline{f(A)}\bigr)$ implies $(\forall B)(B\text{ closed} \implies f^{-1}(B)\text{ closed})$, and that is easily seen (by taking complements) to be equivalent to the continuity of $f$. – Daniel Fischer Nov 12 '16 at 20:10

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