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Prove that function is continuous. Interior.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Suppose $X=A_1 \cup A_2 \cup \ldots ,$ where $A_n \subseteq \text{ Interior of } A_{n+1}$ for each $n$. If $f:X \rightarrow Y$ is a function such that $f|A_n:A_n \rightarrow Y$ is continuous with respect to the induced topology on $A_n$, show that $f$ itself is continuous.

Now here's what I think can be useful in solving the problem.
Let $V$ be open in $Y$. Then $f^{-1}(V)=\bigcup_{i=1}^{i=\infty}{f|A_i}^{-1}(V)$, where each inverse image in the RHS is open in its corresponding restricted space.

I can't see how to go further. Please help.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: prove that function is continuous. Interior.

Suppose $X=A_1 \cup A_2 \cup \ldots ,$ where $A_n \subseteq \text{ Interior of } A_{n+1}$ for each $n$. If $f:X \rightarrow Y$ is a function such that $f|A_n:A_n \rightarrow Y$ is continuous with respect to the induced topology on $A_n$, show that $f$ itself is continuous.

Now here's what I think can be useful in solving the problem.
Let $V$ be open in $Y$. Then $f^{-1}(V)=\bigcup_{i=1}^{i=\infty}{f|A_i}^{-1}(V)$, where each inverse image in the RHS is open in its corresponding restricted space.

I can't see how to go further. Please help.
Let $x\in f^{-1}(V)$. For some $n$, $x\in A_n \subseteq \text{Int}(A_{n+1})$. Then $x\in (f|A_{n+1})^{-1}(V)$, which is open in $A_{n+1}$ (for the induced topology). Thus $(f|A_{n+1})^{-1}(V) = W\cap A_{n+1}$ for some open subset $W$ of $X$. But then $x\in W\cap\text{Int}(A_{n+1})$, which is an open subset of $X$ contained in $f^{-1}(V)$.

That shows that each point of $f^{-1}(V)$ is in the interior of $f^{-1}(V)$, which is therefore an open subset of $X$.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: prove that function is continuous. Interior.

Let $x\in f^{-1}(V)$. For some $n$, $x\in A_n \subseteq \text{Int}(A_{n+1})$. Then $x\in (f|A_{n+1})^{-1}(V)$, which is open in $A_{n+1}$ (for the induced topology). Thus $(f|A_{n+1})^{-1}(V) = W\cap A_{n+1}$ for some open subset $W$ of $X$. But then $x\in W\cap\text{Int}(A_{n+1})$, which is an open subset of $X$ contained in $f^{-1}(V)$.

That shows that each point of $f^{-1}(V)$ is in the interior of $f^{-1}(V)$, which is therefore an open subset of $X$.

thank you so much.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: prove that function is continuous. Interior.

Hello MHB!

I know this thread is really old but somehow I just happened to read it again. I have a question. Can you please see if the following is correct.

Let $X=A_1\cup A_2\cup A_3\cdots$, where $A_i=\text{Int}(A_{i+1})$ for all $i\geq 1$. Let $O$ be open in $A_j$ for some $j$. Then $O=G\cap A_j$ for some $G$ open in $X$. But since $A_j=\text{Int}(A_{j+1})$, $A_j$ itself is open in $X$ and therefore $O$ is open in $X$. So we can say that if a subset $O$ of $X$ is open in one of the $A_i$'s then it is open in $X$ too.

Doesn't the above paragraph provide a simple proof to the problem asked in the first post of this thread?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: prove that function is continuous. Interior.

Hello MHB!

I know this thread is really old but somehow I just happened to read it again. I have a question. Can you please see if the following is correct.

Let $X=A_1\cup A_2\cup A_3\cdots$, where $A_i=\text{Int}(A_{i+1})$ for all $i\geq 1$. Let $O$ be open in $A_j$ for some $j$. Then $O=G\cap A_j$ for some $G$ open in $X$. But since $A_j=\text{Int}(A_{j+1})$, $A_j$ itself is open in $X$ and therefore $O$ is open in $X$. So we can say that if a subset $O$ of $X$ is open in one of the $A_i$'s then it is open in $X$ too.

Doesn't the above paragraph provide a simple proof to the problem asked in the first post of this thread?
I don't see how this can apply to the first post in the thread, because the condition there is $A_n\subseteq\text{Int}(A_{n+1})$, not $A_n=\text{Int}(A_{n+1})$. The sets $A_n$ need not be open.

Here is an interesting example that shows how the result about $f$ being continuous can be false if the hypothesis about the sets $A_n$ is not quite satisfied. Let $X$ be the open unit square in $\mathbb{R}^2$ together with the origin: $X = \{(x,y):0<x<1,\;0<y<1\}\cup\{(0,0\}.$ For $n=1,2,3,\ldots$ let $A_n = \{(x,y)\in X: y\leqslant nx\}.$ Then $X = \bigcup_nA_n.$ also, each point in $A_n$, except for the point (0,0), is in the interior of $A_{n+1}.$ Now define $f$ by $$f(x,y) = \begin{cases} y^2/x& \text{ if }(x,y)\ne(0,0), \\ 0& \text{ if }(x,y)=(0,0).\end{cases}$$ Then $f|A_n$ is continuous for each $n$, but $f$ is not continuous on $X$ because it is not continuous at the origin.
 
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caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: prove that function is continuous. Interior.

I don't see how this can apply to the first post in the thread, because the condition there is $A_n\subseteq\text{Int}(A_{n+1})$, not $A_n=\text{Int}(A_{n+1})$. The sets $A_n$ need not be open.

Here is an interesting example that shows how the result about $f$ being continuous can be false if the hypothesis about the sets $A_n$ is not quite satisfied. Let $X$ be the open unit square in $\mathbb{R}^2$ together with the origin: $X = \{(x,y):0<x<1,\;0<y<1\}\cup\{(0,0\}.$ For $n=1,2,3,\ldots$ let $A_n = \{(x,y)\in X: y\leqslant nx\}.$ Then $X = \bigcup_nA_n.$ also, each point in $A_n$, except for the point (0,0), is in the interior of $A_{n+1}.$ Now define $f$ by $$f(x,y) = \begin{cases} y^2/x& \text{ if }(x,y)\ne(0,0), \\ 0& \text{ if }(x,y)=(0,0).\end{cases}$$ Then $f|A_n$ is continuous for each $n$, but $f$ is not continuous on $X$ because it is not continuous at the origin.
Thanks so much!