# Identity map between metric spaces continuous or not. [closed]

How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?

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• You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)\lt\epsilon$. – John Douma Dec 21 '18 at 4:03
• Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help. – Matt A Pelto Dec 21 '18 at 4:29

a. Let $$\varepsilon>0$$ be given. Since $$\sup_{x\in [0,1]} | \, f(x)-g(x)| \geq \int_0^1 |\, f(x)-g(x)|dx$$, we may choose $$\delta=\varepsilon$$ so that $$\int_0^1 |\, f(x)-g(x)|dx<\varepsilon$$ whenever $$\sup_{x\in [0,1]} |\, f(x)-g(x)|<\delta$$. Therefore $$id:X_1 \longrightarrow X_2$$ is continuous.

b. Notice $$\int_0^1 |x^n|dx \to 0$$ as $$n \to \infty$$ BUT $$\sup_{x\in [0,1]} |x^n|=1$$ for every $$n \in \mathbb N$$. Therefore $$id:X_2 \longrightarrow X_1$$ is NOT continuous at the constant function $$g \equiv 0$$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).

c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.

Technically for b. I used the fact that the functions $$f_n(x):=x^n$$ from the sequence of functions $$\{f_n\}_{n=1}^\infty$$ converge to the constant function $$g \equiv 0$$ in $$(X_2, d_2)$$ but the same is not true in $$(X_1, d_1)$$. And so $$id: X_2 \longrightarrow X_1$$ is not sequentially continuous at $$g$$ as opposed to identifying the pointwise limit $$f(x):=\lim_{n\to \infty} f_n(x)$$ which is not in the set $$\mathcal C[0,1]$$ as $$f$$ is not continuous at $$x=1$$.

If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $$[0,1]$$ (Cauchy in $$(X_1, d_1)$$) avoids this. For this argument we make the following two observations, $$|x^n-x^m|=|x^n||1-x^{m-n}|$$ and the function $$F(x):=\left(1-\frac1x\right)^x$$ is nondecreasing on $$[1, \infty)$$ with $$\lim_{x \to \infty} F(x)=\frac1e$$. So with $$\varepsilon=\frac1{4}(1-e^{-1})$$ and for any $$n\geq 2$$, we may select $$m=2n$$ and have $$\sup_{x\in[0,1]} |x^n-x^m| \geq \left(1-\frac1n\right)^n\left(1-\left(1-\frac1n\right)^n\right)\geq \varepsilon$$ which shows that the sequence of functions $$\{f_n\}_{n=1}^\infty$$ is not Cauchy in $$(X_1, d_1)$$ by definition.

• For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous. – ChakSayantan Dec 21 '18 at 6:09
• You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump). – Matt A Pelto Dec 21 '18 at 9:43
• Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs. – Matt A Pelto Dec 21 '18 at 10:15
• Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)\leq K\cdot d_1(f,g)$ for all $f,g\in X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $id_X:(X,d_1)\to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$ – DanielWainfleet Dec 22 '18 at 3:06
• Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_- – Matt A Pelto Dec 22 '18 at 3:24

We know that a continuous map between metric spaces is one which preserves limits: $$f$$ is continuous if and only if $$\lim\limits_{n\to\infty} f(x_n)=f(\lim\limits_{n\to\infty} x_n)$$ for each convergent sequence $$x_n$$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.

For example, I don't think that b) is true. Consider the sequence $$f_n(x)=x^n.$$

Then $$\lim\limits_{n\to\infty} f_n=0$$ in the metric $$d_2$$, but not in $$d_1$$: it even fails to be Cauchy in $$d_1$$.

Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $$x\in X_i$$, $$d_j(x,0)\le M\cdot d_i(x,0)$$ -- then the identity map from $$X_i$$ to $$X_j$$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.

• nice counterexample – Matt A Pelto Dec 21 '18 at 4:25