1
$\begingroup$

I am trying to solve this exercise in Tao's analysis text.

Let $X$ be a subset of $\mathbf{R}$ be a continuous function. If $Y$ is a subset of $X$, show that the restriction $f \mid_Y : Y \to \mathbf{R}$ of $f$ to $Y$ is also a continuous function.

Here is my attempt at a proof.

Define the inclusion map $i: Y \to X$. Since $f$ is continuous by assumption, we have \begin{align*} \forall \epsilon > 0, \forall x_0 \in X, \exists \delta > 0, \forall x \in X, |x - x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon. \end{align*} We first show that $i$ is continuous on its entire domain. Let $\epsilon > 0$. Pick an arbitrary $y_0 \in Y$. Then, by continuity of $f$, there exists $\delta$ such that for any $y \in Y$ (which is also in $X$), $|y - y_0| < \delta$ implies $|i(y) - i(y_0)| < \epsilon$. Thus, $i$ is continuous. Since the composition of continuous functions is continuous, $f \circ i$ is continuous. but $f \mid_Y = f \circ i$. Thus, the restriction is also continuous.

How does this look? I worry the proof could be lacking rigor, particularly in explaining why $i$ is continuous. All I am trying to say is that the same $\delta$ from continuity of $f$ should work for $i$ since "$\forall x \in X$" in the definition of continuity of $f$ certainly applies for a subset of $X$.

$\endgroup$
0
$\begingroup$

To prove the continuity of the function $i $, you should show the existence of a $\delta \gt0$ such that the definition of continuity holds. I don't see how the continuity of $f $ guarantees such a $\delta$ for the continuity of $i $.

For the inclusion map, the obvious choice is to take $\delta =\varepsilon$. Apart from that, your proof looks fine.

$\endgroup$
3
  • $\begingroup$ Thank you for the response. Could you explain a bit more why $\delta = \epsilon$ works? I have tried this, and it seems I would need to prove that $|y - y_0| < \delta = \epsilon$ implies that $|i(y) - i(y_0)| = |x - x_0| < \epsilon$. I am not sure how his follows. It also doesn't seem to use continuity of $f$ (though perhaps that fact isn't needed). $\endgroup$ – user465188 Jun 23 '19 at 3:33
  • 1
    $\begingroup$ $i (y)=y $ for all $y\in Y $. Inclusion map is the identity map restricted to $Y $. $\endgroup$ – Shivering Soldier Jun 23 '19 at 3:36
  • 1
    $\begingroup$ Thank you! This was very helpful. $\endgroup$ – user465188 Jun 23 '19 at 3:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy