Welcome to our community

Be a part of something great, join today!

Continuity stuff

Megus

New member
Oct 18, 2012
4
If $f: (M_1,d_1)\to(M_2,d_2)$ is continuous and $N_1$ is a vector subspace of $M_1$ then $f\bigg|_{N_1}(N_1,d_1)\to (M_2,d_2)$ is continuous.

How does one prove continuity with $f$ restricted into a set?

Let $f: (M_1,d_1)\to (M_2,d_2)$ be continuous and $f(M_1)\subseteq N_2\subseteq M_2$ then $f: (M_1,d_1)\to(N_2,d_2)$ is continuous.

I need a hint for this one.

Let $(M_1,d_1),\ldots,(M_n,d_n),$ and $M=M_1\times\cdots\times M_n.$ Prove the following:
$f: (N,d)\to (M,d_e)$ is continuous iff $f_i: (N,d)\to(M_i,d_i)$ is continuous for $1\le i\le n.$

The left implication is easy because if each $f_i$ is continuous then clearly $f$ is continuous, but don't know how to prove the right implication.

Thanks!
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Are you sure you mean vector subspace? Perhaps it is topological subspace.

Well, if $f$ is continuous, it means that for all $x,x_0 \in M_1$ and $\varepsilon >0$, we have that exists $\delta >0$ such that $d_1(x,x_0) < \delta$ implies $d_2(f(x),f(x_0)) < \varepsilon$.

Since this is valid for all $M_1$, then it is valid for $N_1$ since $N_1 \subset M_1$, therefore all $x \in N_1$ also belong to $M_1$. Consequently, for all $x, x_0 \in N_1$ and $\varepsilon >0$, we have that exists $\delta >0$ such that $d_1(x,x_0) < \delta$ implies $d_2(f(x),f(x_0)) < \varepsilon$.

The second and third questions seem to follow the same idea. How about you try them now? :D
 

Megus

New member
Oct 18, 2012
4
Okay, the second one struggles me because of the $f(M_1)\subseteq N_2,$ I don't know how to use it, can you enlight me?

Thanks.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
The argument goes similarly to what I did for the first part, let's see how. (Nod)

If $f: (M_1, d_1) \to (M_2, d_2)$ is continuous, then for all $x, x_0 \in M_1$ and $\varepsilon >0$, we have that exists $\delta >0$ such that $d_1(x,x_0) < \delta$ implies $d_2(f(x),f(x_0)) < \varepsilon$.

The set of all images of $f$ is denoted by $f(M_1)$ and it's a subset of $M_2$, therefore all $f(x) \in M_2$. Using the hypothesis that $f(M_1) \subseteq N_2$, we have that all $f(x) \in N_2$ too.

Repeating it all together, for all $x, x_0 \in M_1$ and $\varepsilon >0$, we have that exists $\delta >0$ such that $d_1(x,x_0) < \delta$ implies $d_2(f(x), f(x_0)) < \varepsilon$, where $f: (M_1, d_1) \to (N_2, d_2)$.

This proves it is continuous. ;)