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Problem of the Week #11 - August 13th, 2012

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Chris L T521

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Jan 26, 2012
995
Many thanks to Opalg for suggesting this week's problem.

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Problem: Consider the following:

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.
Show that the above result is false if the words "Interior of" are removed. Specifically, give an example of a topological space $X$, an increasing nest of subspaces $A_n$ with $X=A_1 \cup A_2 \cup \ldots ,$ and a function $f:X \rightarrow Y$ such that, for each $n$, $f|A_n:A_n \rightarrow Y$ is continuous with respect to the induced topology on $A_n$, but such that $f$ is not continuous on $X$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
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Jan 26, 2012
995
No one took a shot at this problem. Here's the solution.

Take $X$ to be the right-hand half-plane together with 0: $X = \{(x,y)\in\mathbb{R}^2:x>0\}\cup\{(0,0)\}$. For each $n$, let $A_n = \{(x,y)\in X:|y| \leqslant nx.$ Define $f: X\to\mathbb{R}$ by $$f(x,y) = \begin{cases}y^2/x & \text{if }(x,y)\ne(0,0), \\ 0 & \text{if }(x,y) = (0,0). \end{cases}$$

Clearly $f$ is continuous everywhere except perhaps at the origin. If $(0,0)\ne(x,y)\in A_n$ and $\|(x,y)\| <\varepsilon/n^2,$ then $|f(x,y)| = \dfrac{y^2}x \leqslant \dfrac{n^2x^2}x = n^2x <\varepsilon.$ Thus $\displaystyle \lim_{(x,y)\to(0,0)}f(x,y) = 0 = f(0,0).$ Hence $f|A_n$ is continuous.

But $(1/k^2,1/k)\in X$ for $k = 1,2,\ldots,$ $(1/k^2,1/k) \to (0,0)$ and $f(1/k^2,1/k) = 1$. Thus $f$ is not continuous at $(0,0).$
 
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