If $E_i$ is a topological space and $Ω_i\subseteq E_i$, how does continuity of $f:Ω_1\toΩ_2$ at $x_1\inΩ_1$ belong on the chosen topologies?

Question

Let

• $(E_i,\tau_{E_i})$ be a topological space
• $\Omega_i\subseteq E_i$ and $\tau^{E_i}_{\Omega_i}$ denote the subspace topology on $\Omega_i$ inherited from $E_i$
• $f:\Omega_1\to E_2$ with $f(\Omega_1)\subseteq\Omega_2$
• $x_1\in\Omega_1$

Consider the following statements:

1. $f$ is $\left(\tau_{E_1},\tau_{E_2}\right)$-continous at $x_1$
2. $f$ is $\left(\tau_{\Omega_1}^{E_1},\tau_{E_2}\right)$-continous at $x_1$
3. $f$ is $\left(\tau_{E_1},\tau_{\Omega_2}^{E_2}\right)$-continous at $x_1$
4. $f$ is $\left(\tau_{\Omega_1}^{E_1},\tau_{\Omega_2}^{E_2}\right)$-continous at $x_1$

What are the implications between (1.) to (4.)? Are they all equivalent?

I was able to show that (1.) implies (2.). Now, let's for example assume (2.) and let $N_2$ be a $\tau_{E_2}$-open neighborhood of $f(x_1)$. By $(2.)$, there is a $\tau_{\Omega_1}^{E_1}$-open neighborhood $\tilde N_1$ of $x_1$ with $$f(\tilde N_1)\subseteq N_2.$$ Since $\tilde N_1\in\tau_{\Omega_1}^{E_1}$, there is a $N_1\in\tau_{E_1}$ with $$\tilde N_1=\Omega_1\cap N_1.$$ And since $x_1\in\tilde N_1$, we've got $x_1\in N_1$. However, since $N_1\not\subseteq\tilde N_2$, we're not able to conclude $$f(N_1)\subseteq f(\tilde N_1)\subseteq N_2.$$

The point above is that we could assume $N_1\subseteq\tilde N_1$ (i.e. $N_1\subseteq\Omega_1$), if $\Omega_1$ would belong to $\tau_{E_1}$. So, do we need to assume that $\Omega_1$ is $\tau_{E_1}$-open?

Answer 1

(i). $1\iff 3$.... (ii). $2\iff 4$.... (iii). $1\implies 2$.... (iv). $2$ does not imply $1$.

For (i) and for (ii): Let $f:A\to B.$ This literally means that $f$ maps $A$ into $B.$ Let $B$ be a subspace of $C.$ Then we also have $f:A\to C.$

Now $f:A\to B$ is continuous iff $f:A\to C$ is continuous. PROOF:

(I). If $f:A\to C$ is continuous: Let $U$ be any open subset of the space $B,$ so $B=V\cap B$ for some open subset $V$ of the space $C.$ And $f^{-1}V$ is an open subset of $A,$ because $f:A\to C$ is continuous. But $f$ maps $A$ into $ B$ so $$f^{-1}U=f^{-1}(V\cap B)=\{p\in A: f(p)\in V\land f(p)\in B\}=$$ $$=\{p\in A: f(p)\in V\}=f^{-1}V.$$ So $f^{-1}U=f^{-1}V$, which is open in $A.$

(II). If $f:A\to B$ is continuous: Let $V$ be any open subset of $C.$ Since $f$ maps $A$ into $B$ we have $$f^{-1}V=\{p\in A: f(p)\in V\}=\{p\in A: f(p)\in V\land f(p)\in B\}=$$ $$=\{p\in A:f(p)\in V\cap A\}=$$ $$=f^{-1}(V\cap A).$$ Now $V \cap B$ is an open subset of $B,$ and $f:A\to B$ is continuous. So $f^{-1}V=f^{-1}(V\cap B),$ which is open in A.

For (iii). For $f:A\to B,$ the notation $f|_D$ means the restriction of $f$ to the domain $ D$.

If $f:A\to B$ is continuous and $D$ is a subspace of $A$ then $f|_D:D\to B$ is continuous. PROOF:

Let $U$ be any open subset of $B.$ We have $$(f|_D)^{-1}U= \{p\in D: f(p)\in U\}=D \cap \{p\in A:f(p)\in U\}=D \cap f^{-1}U.$$ Now $f:A\to B$ is continuous so $f^{-1}U$ is an open subset of $A$. So $(f|_D)^{-1}U=D\cap f^{-1}U$ is an open subset of $D$ by the def'n of subspace.

For (iv). If $D$ is a subspace of $A$ and if $f:A\to B$ such that $f|_D$ is continuous, it does not follow that $f:A\to B$ is continuous. For example let $A=(0,1]$ and $D=(0,1),$ and let $B=\Bbb R,$with the usual topologies on $A$ and on $\Bbb R.$ Let $f(x)=0$ for $x\in D$ and let $f(1)=1.$

Remark: $f_D$ can also be written as $f|D.$

Answer 2

You consider the three functions $f : E_1 \to E_2$, $f' : \Omega_1 \to E_2$ and $f'' : \Omega_1 \to \Omega_2$, where $\Omega_i$ is understood as a subspace of $E_i$. If $f$ is continuous at $x_1$, then $f'$ and $f''$ are continuous at $x_1$ as they are restrictions of $f$. Moreover, $f'$ is continuous at $x_1$ if and only $f''$ is. However, if $f'$ is continuous at $x_1$, then in general $f$ is not continuous at $x_1$ (consider e.g. $\Omega_1 = \lbrace x_1 \rbrace$; in this case $f'$ is always continuous). Finally, item 3. doesn't make sense because you you don't have a function defined on $E_1$ with values in $\Omega_2$.