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Definition 4.5

Suppose $X$ and $Y$ are metric spaces, $E \subset X, p \in E$ and $f$ maps $E$ into $Y$. Then $f$ is said to be continuous at $p$ if for every $\varepsilon > 0$ there exists a $\delta >0$ such that $$d_Y(f(x),f(p)) < \varepsilon$$ for all points $x \in E$ for which $d_X(x,p) < \delta$.

Theorem 4.6

In the situation given in Definition 4.5, assume also that $p$ is a limit point of $E$. Then $f$ is continuous at $p$ if and only if $\lim_{x \to p} f(x) = f(p)$.

Rudin didn't write Definition 4.5 as follows:

Definition 4.5'

Suppose $X$ and $Y$ are metric spaces, $p \in X$ and $f$ maps $X$ into $Y$. Then $f$ is said to be continuous at $p$ if for every $\varepsilon > 0$ there exists a $\delta >0$ such that $$d_Y(f(x),f(p)) < \varepsilon$$ for all points $x$ for which $d_X(x,p) < \delta$.

And Rudin didn't write Theorem 4.6 as follows:

Theorem 4.6'

In the situation given in Definition 4.5', assume also that $p$ is a limit point of $X$. Then $f$ is continuous at $p$ if and only if $\lim_{x \to p} f(x) = f(p)$.

And Rudin wrote Theorem 4.8 as follows:

Theorem 4.8

A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$.

Why?

Why is $E$ necessary in Definition 4.5, Theorem 4.6, Theorem 4.7?

I think $E$ is redundant.

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    $\begingroup$ If you take $(E, d|_{E\times E}) $ as the metric subspace, then you can see the two defintions are in fact the same. $\endgroup$ – lEm Jan 31 at 11:14
  • $\begingroup$ Thank you, lEm for your comment. $\endgroup$ – tchappy ha Jan 31 at 13:23
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(Rudin's related explanation in the later chapter)

You are right here. He can simply discard the complement of $E$ in $X$, and view it as a function from $E$ to $Y$. As IEm points out, E is a metric space in its own right using the metric inherited from $X$.

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  • $\begingroup$ Thank you very very much, YuiTo Cheng for the information. $\endgroup$ – tchappy ha Jan 31 at 13:22
  • $\begingroup$ This raises a meta-question: what was Rudin thinking by including $E$ and a remark that it's useless? He doesn't strike me as one to include fluff. $\endgroup$ – Solomonoff's Secret Jan 31 at 15:24
  • $\begingroup$ @Solomonoff'sSecret Not actually... For example, his proof are sometimes unnecessarily longer than the standard proof. $\endgroup$ – YuiTo Cheng Jan 31 at 15:28

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