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Two disjoint sets $A$ and $B$, neither empty, are said to be mutually separated if neither contains a boundary point of the other. A set is disconnected if it is the union of separated subsets, and is called connected if it is not disconnected.

With the above definition of connected set, how to prove that the set $A=\{(x, y)\in \mathbb{R}^2:|x|=|y|\}$ is connected.

Exercise should try using the definition of previous related and for this I guess I should do the proof by contradiction but do not know how to proceed.

Roiner Segura Cubero
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Do you know straight lines are connected set in $\mathbb{R}^2$? do you know union of connected set is connected if they have one point in common?

Now can you see your set with above stated properties?

Myshkin
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You can prove that $\{y=x\}$ and $\{y=-x\}$ are connected. Since they have a point in common, their union is. Can you try to argue this is the same, in essence, than showing the real line is connected?

First, let's obtain a slightly more useful equivalent to your definition. First, if $A,B$ are open sets and $A\cap B=\varnothing$, then they are separated. Now suppose $A,B$ are separated sets, and $C=A\cup B$. Then $C\setminus \bar A$ and $C\setminus \bar B$ are relatively open in $C$. But $$(A\cup B)\setminus \bar A=B$$ since $B\cap \bar A=\varnothing$. Thus $C=A\cup B$ and $A,B$ are relatively open in $C$. We can then just say that a set $C$ is disconnected if it can be written as the disjoint union of relatively open subsets of $C$.

Now, suppose $f:C\to C'$ is continuous and onto. If $C'$ is disconnected by $A,B$, then $C$ is disconnected by the open sets $f^{-1}(A)$, $f^{-1}(B)$. Thus, by the contrapositive, if $C$ is connected, so must be $C'$. That is, the image of a connected set under a continuous function is connected.

Now consider the function $f:\Bbb R\to\Bbb R^2$ given by $f(x)=(x,x)$. Then your set is $f(\Bbb R)$. I hope you know that $\Bbb R$ is connected. You should see $f$ is continuous: $$\lVert f(x)-f(y)\rVert^2=2(x-y)^2$$

so if $x$ and $y$ are close, so is $(x,x)$ and $(y,y)$. Thus, $f(\Bbb R)$ is connected.

It is also true that the union of connected sets with non-empty interesction is connected. And one can prove it as follows. First, we prove that $C$ is connected if and only if the only continuous functions $f:C\to\{0,1\}$ $-$ where $\{0,1\}$ is given the discrete metric (topology), that is, all sets are open $-$ are constant.

If $C=A\cup B$ with $A,B$ open, then we can define $f(a)=1$ when $a\in A$ and $f(b)=0$ when $b\in B$, to get non constant continuous function $f:C\to\{0,1\}$. Conversely, if we had a non-constant continuous function $f:C\to\{0,1\}$, we could write $C=f^{-1}(\{0\})\cup f^{-1}(\{1\})$ where this two sets are open and disjoint.

With this out of the way, suppose that we have a collection $\mathscr C$ of connected sets, and they have a point in common, call it $c$, that is $\bigcap\mathscr C\neq \varnothing$. Consider a continuous function $f:\bigcup \mathscr C\to \{0,1\}$. We will show that $f(x)=f(c)$ for any $x\in\bigcup\mathscr C$. Indeed, pick $x$ in the union. Then this belongs to some of the sets in the collection, call this set $C_x$. Then $f$ as a function from $C_x$ to $\{0,1\}$ is still continuous, so, since $C_x$ is connected, it must be constant. But $c\in C_x$; so we must have $f(x)=f(c)$. Since this $x$ was arbitrary, the claim follows.

Pedro
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  • Tamaroff...how to prove that $A={(x, y)\in \mathbb{R}^2:y=x}$ is connected? – Roiner Segura Cubero Aug 16 '13 at 04:53
  • @RoinerSeguraCubero One usually shows that continuous functions map connected sets to connected sets. – Pedro Aug 16 '13 at 05:03
  • Peter in the course have not seen my teacher continuity of functions you want the test to be performed only using the definition of connectedness given – Roiner Segura Cubero Aug 16 '13 at 05:19
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    @RoinerSeguraCubero Are you really studying connectedness without knowing about continuity...? – Pedro Aug 16 '13 at 05:23
  • yes...is that what you want is to do the exercise only with the definition ... I know of a related image under a continuous function is connected and parameterization can prove that a segment is connected ... but I wonder if the exercise can be performed using only the definition of related, perhaps through a contradiction – Roiner Segura Cubero Aug 16 '13 at 05:30
  • @RoinerSeguraCubero I guess it would be very tortous. I'd rather develop the machinery to make the proofs easy! Cf. my edit. – Pedro Aug 16 '13 at 05:31
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If |x| < |y| then those x's are all in the disk of radius y about the origin. That is surely connected

Betty Mock
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    Betty sorry the set is $A={(x, y)\in \mathbb{R}^2:|x|=|y|}$ – Roiner Segura Cubero Aug 16 '13 at 04:49
  • $\Bbb Q\cap[0,1]$ is a subset of $[0,1]$ which is connected, but the rationals are surely not connected. – Pedro Aug 16 '13 at 05:41
  • @Roiner -- don't know how I misread that -- my version of the problem is easier anyway. B – Betty Mock Aug 17 '13 at 03:25
  • -1: This is not written carefully enough for me to easily perceive its meaning, but it certainly does not answer the OP's question, i.e., give a proof of the connectedness of the locus $y = \pm x$ in $\mathbb{R}^2$. – Pete L. Clark Aug 17 '13 at 10:34