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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?

Suppose $T$ is uniformly continuous on $M$, and let $p$ be an arbitrary point of $M$. So, for every real $\varepsilon > 0$, there exists $\delta > 0$ such that $$d_Y\left( T(x), T(p) \right) < \varepsilon$$ for all points $x \in M$ for which $$d_X\left( x, p \right) < \delta.$$

Thus the restriction $T \vert_M$ of the mapping $T$ to set $M$ is continuous at point $p$ and hence continuous on the set $M$.

How to show from here that $T$ is continuous at point $p$?

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  • $\begingroup$ Please check your definition of uniform continuity. The condition does not depend on $M$. $\endgroup$ – gerw Sep 29 '16 at 18:18
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The answer is yes and there's nothing to show really: Note that the choice of $\delta$ in the definition of uniform continuity depends on $\epsilon$ only and hence satisfies the definition of pointwise continuity at any point at the same time.

In other words, in the definition of continuity $\delta$ is a function of both $p$ and $\epsilon$; uniform continuity is the special case, where $\delta$ can be chosen to be constant in $p$, hence is a function of $\epsilon$ only.

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  • $\begingroup$ so nice of you! But I'd really appreciate if you could also answer my question specifically in its context. That is, given an arbitrary but fixed point $p \in M$, how do we guarantee that, $$d_Y\left( T(x), T(p) \right) < \varepsilon$$ for all points $x \in X$ for which $$d_X(x,p) < \delta,$$ given that $$d_Y\left( T(x), T(p) \right) < \varepsilon$$ for all points $x \in M$ for which $$d_X(x,p) < \delta?$$ Is my question a valid one? If so, I'd be really grateful for your answer! $\endgroup$ – Saaqib Mahmood Oct 1 '16 at 16:31
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If given an epsilon, you can pick a delta ball in the domain that works for any two values $x_1,x_2$ such that the distance between the values of your function in the range are bounded by epsilon, can you do the same for that epsilon and some particular pair of points in the domain?

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  • $\begingroup$ thank you for your answer, but can you please have a look at my comment upon Damian Reding's answer and then see if you could also answer my specific query? $\endgroup$ – Saaqib Mahmood Oct 1 '16 at 16:34

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