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I'm working on a topology homework problem and a hint would be appreciated. The idea is, we have a connected subspace $A$ of $\mathbb{R}^{2}$ and we define $A_{\varepsilon}$ to be the set of all points in $\mathbb{R}^{2}$ that are within $\varepsilon$ of some point in $A$, and we want to prove $A_{\varepsilon}$ is path connected.

$A_{\varepsilon}$ is open in $\mathbb{R}^{2}$, so it suffices to show it is connected. The way I've been trying to go about this is by showing if $A_{\varepsilon}$ isn't connected, then $A$ isn't connected (which we've assumed it is). So suppose $A_{\varepsilon}$ isn't connected, take some separation $U, V$, and use it to produce a separation $U$ intersect $A$, $V$ intersect $A$ of $A$. That works great if $U$ and $V$ both contain points in $A$. But what if, say, $U$ contains all of $A$, and $V$ contains none of $A$? Then our separation as defined above isn't a separation because one of the sets will be empty. Is it possible this can happen? Intuitively it feels to me like somehow if $V$ has no points in $A$ then the points in $V$ are somehow "too far" to have been in $A_{\varepsilon}$. But I'm getting kind of lost in the weeds here. Can anyone tell me if my approach is salvageable?

David G. Stork
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Peter A
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1 Answers1

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Your approach can indeed be made to work with a little more thought. However, thinking about how to complete your argument made me realize that there's an easier approach altogether:

For every $a\in A$, define $S_a = A \cup a_\varepsilon$, where $a_\varepsilon$ is the set of all points within $\varepsilon$ of $a$. Then:

  • $A_\varepsilon = \bigcup_{a\in A} S_a$;
  • $\bigcap_{a\in A} S_a \supset A$ and thus is nonempty (we may assume $A$ is nonempty);
  • each $S_a$, being the union of two connected sets $A \cup a_\varepsilon$ with nonempty intersection (it contains $a$), is itself connected.

Therefore $A_\varepsilon$, being the union of a family of connected sets $S_a$ whose intersection is nonempty, is itself connected.

(This is similar to the proof that if $U$ and $V$ are connected then $U\times V$ is connected.)

Greg Martin
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