1

Question:

Let f and g be continuous mappings of a metric space X into a metric space Y and let E be a dense subset of X. Prove that f(E) is dense in f(X). If $f(p) = g(p)$ for all p $\in$ E prove that $f(p) = g(p)$ for all X.

Answer:

  1. Let's show that every point of f(X) is either a point of f(E) or a limit point of f(E) or both. Suppose that y $\in$ f(X). Then, there exists a point p $\in$ X such that f(p) = y. Since E is dense in X, p $\in$ E U E$'$. If p $\in$ E then y $\in$ f(E) and we are done. If p $\in$ E$'$ and p is not in E, then there exists a sequence p$_n$ $\in$ E such that p$_n$ -> p. We then have f(p$_n$) -> f(p). That is , f(p) $\in$ f(E).
  2. Now, suppose f(p) = g(p) for all p $\in$ E. Let x $\in$ X\E. Since E is dense in X we have a sequence q$_n$ $\in$ E such that q$_n$ -> x. So, f(x) = f(lim q$_n$) = lim f(q$_n$) = lim g(q$_n$) = g(lim q$_n$) = g(x). Thus, f(x) = g(x) for all x $\in$ X

Could you tell me if the bold line is legitimate and explain me why it is legitimate if it is? Thank you so much!

Gigili
  • 5,503
  • 8
  • 41
  • 62
InfimumMaximum
  • 1,406

1 Answers1

2

Continuity for topological spaces (and metric spaces) means that every open set has an open preimage. Convergence means that for every neighborhood of the limit point, the sequence eventually lies entirely in the neighborhood.

To be clear, by "neighborhood of a point" I mean any open set containing that point.

So, to see that whenever $p_n\to p$, $f(p_n)\to f(p)$, consider an arbitrary neighborhood $V$ of $f(p)$. This neighborhood has an open set as a preimage (as $f$ is continuous) containing $p$, so it will be a neighborhood, $U$, of $p$. Then there is some $N$ such that $p_n\in U$ whenever $n>N$, so that $f(p_n)\in V$ whenever $n>N$. As $V$ was arbitrary, this completes the proof.

Note that this works for any topological spaces, not just metric spaces.

Andrew Poelstra
  • 575