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Let $\left(X, d_X \right)$ and $\left(Y, d_Y \right)$ be metric spaces, let $M$ be a subset of $X$, and let $T \colon X \to Y$ be a mapping that is uniformly continuous on $M$. Then is $T$ also continuous on $M$?

Definition of Continuity:

Let $\left(X, d_X \right)$ and $\left(Y, d_Y \right)$ be metric spaces, let $T \colon X \to Y$ be a mapping, let $M$ be a subset of $X$, and let $p$ be a point of $X$. Then $T$ is said to be continuous at point $p$ if, for every real numnber $\varepsilon > 0$, we can find a real number $\delta > 0$ such that $$d_Y\left( T(x), T(p) \right) < \varepsilon$$ for all points $x \in X$ for which $$ d_X(x, p) < \delta.$$

If $T$ is continuous at each point $p \in M$, then $T$ is said to be continuous on set $M$.

Finally, if $T$ is continuous on $X$, then $T$ is said to be continuous.

Definition of Uniform Continuity:

Let $\left(X, d_X \right)$ and $\left(Y, d_Y \right)$ be metric spaces, let $T \colon X \to Y$ be a mapping, and let $M$ be a subset of $X$. Then $T$ is said to be uniformly continuous on $M$ if, for every real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that $$d_Y\left( T(x_1), T(x_2) \right) < \varepsilon$$ for all points $x_1, x_2 \in X$ for which $$d_X\left( x_1, x_2 \right) < \delta.$$

Are these two sets of definitions correct?

Further i want to prove the inclusion as mentioned above, only by definition, but unable to get it rightly.

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  • $\begingroup$ The definitions are correct. Let $T \colon X → Y$ be uniformly continuous. Let $ε > 0$ be a real number and $p ∈ X$ be any point. Now, you need to find some real $δ > 0$ such that for all $x ∈ X$, $d_X(x,p) < δ ⇒ δ_Y(T(x),T(p)) < ε$. Any ideas what $δ$ you may pick? $\endgroup$
    – k.stm
    May 22, 2017 at 21:05
  • $\begingroup$ If $\delta$ is $\epsilon/2$? $\endgroup$
    – Fernando
    May 22, 2017 at 21:07
  • $\begingroup$ No. You have to use the fact that $T$ is uniformly continuous. Can you see how? $\endgroup$
    – k.stm
    May 22, 2017 at 21:08
  • $\begingroup$ yeah sure, it would be helpful if you elaborate $\endgroup$
    – Fernando
    May 22, 2017 at 21:09
  • $\begingroup$ Well, you got some $ε > 0$ and need to find some $δ > 0$ (with some properties). How may these two relate to the definition of uniformly continuity (as fulfilled by $T$)? $\endgroup$
    – k.stm
    May 22, 2017 at 21:12

2 Answers 2

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Here is how I think about it, without too many calculations:

Continuity means that for any $\epsilon$ and any $x$ it must be possible to find a $\delta$ that works. Uniform continuity means that for any $\epsilon$ it's possible to find a single $\delta$ that works for all $x$ simultaneously. Clearly, if you're given an $\epsilon$ and an $x$, and you know that can find a $\delta$ that works for all $x$ simultaneously, then you can find one that works for your specific $x$.

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There is no way I am saying anything different from Aurthur, but, maybe this would explain it:

$T$ is said to be continuous at point $p$ if, for every real numnber $\varepsilon > 0$, we can find a real number $\delta(p) > 0$ such that $$d_Y\left( T(x), T(p) \right) < \varepsilon$$ for all points $x \in X$ for which $$ d_X(x, p) < \delta(p).$$

What this means is that for continuous functions all you need is to find such a $\delta(p)$ that would work at the point p, and your choice for $\delta$ can depend on the point at which you are proving continuity.

Now, uniform continuity is a much stronger condition, as it requires a one size fit all $\delta$. It says that $T$ is uniformly continuous on $M$ if, for every real number $\varepsilon > 0$, we can find a real number $\delta^* > 0$ such that $$d_Y\left( T(x_1), T(x_2) \right) < \varepsilon$$ for all points $x_1, x_2 \in X$ for which $$d_X\left( x_1, x_2 \right) < \delta^*.$$

What this means is that for all $p\in M$, once we put $p=x_1$ and $\delta(p)=\delta^*$ from right above, our job of finding the small enough $\delta$ required for continuity at p is done! (Note to you: Look at where the quantifiers lie)

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