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Question: If f and g are continuous mappings of a metric space X into a metric space Y, let E be a dense subset of X. if g(p) = f(p) for all p $\in$ E, prove that g(p)= f(p) for all p$\in$ X.

Answer: Now, suppose f(p) = g(p) for all p ∈ E. Let x ∈ X\E. Since E is dense in X we have a sequence q$_n$ ∈ 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 ∈ X

Hi, I founded the above answer and I'm not sure if the bold line is true and how so if it is. Could you help me understand the equalities in the bold line?

Thank you!

Martin Sleziak
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InfimumMaximum
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2 Answers2

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The bold line is true because of continuity of $f$ and $g$. This is because sequential continuity is equivalent to metric continuity. That is, the $\epsilon-\delta$ definition of pointwise continuity of a function $f$ is the same as $\lim_{n\to\infty}f(x_{n})=f(x)$ for all sequences $(x_{n})_{n\in\mathbb{N}}$ that converge to $x$.

user71352
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$$f(\lim q_n) = \lim f(q_n) = \lim g(q_n) = g(\lim q_n)$$

We can switch limit and function (equalities 1 and 3) by continuity of $f$ and $g$. Equality 2 is true because $f(q_n)$ and $g(q_n)$ are two sequences which are equal in every entry. By continuity of $f$ and $g$, they both have limits; since they are equal, their limits must be the same.

Eric Auld
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