i need help with this question. i know path connected => connected.
i) i dont know that this graph looks like but i dont think it is path conected.
ii) this is not path connected
iii) i dont think is path connected
i have no idea with connected.

am i on the right track? will pay whoever can help, thnks

i) Is path-connected.
It is the union of the graphs :
y = cos x
y = - cos x

that both represent continuous functions, and show some common points (for example x = pi/2 and y = 0).

ii) Is not connected (hence not path-connected).
Let U = R+* x R and V = R-* x R.
U and V are both open sets, they are disjoint, R² - Q² is included into the union U υ V, and both U and V meet R² - Q².

iii) Should be connected but not path-connected (one example of each case )
Let check this true :

It is clearly not path-connected, since if you start from any point on the Y-axis, it's impossible to connect it to any given point of the form y = cos (1/x) with a continuous path.

Let show that it is connected :
Let A the set of iii), if it was not the case, you would have two open sets U and V such that :
A ⊆ U υ V
U ∩ V = ∅ (U and V are disjoint)
U ∩ A ≠ ∅ (U meets A)
V ∩ A ≠ ∅ (V meets A)

Take a point M of the Y-axis, of y-coordinate between -1 and 1. It lies into A, so it lies into U or into V. We may assume it lies into U.
Since the Y-axis is path-connected, all the Y-axis is contained into U.
Since U is an open set and M ∈ U, U contains an open ball around M. But for any radius of such open ball, it will meet the graph : y = cos(1/x).
Hence, that open ball (and U) contains a point of that graph.
And since that graph is path-connected, all that graph is included into U.
To sum up, A ⊆ U, which is impossible since, in this case, V should be empty !

Hope it is clear enough, pm if needed.

I just quickly glanced at this post. Just a correction in feanur's answer for ii) this subset is actually path connected and hence is connected. The reason U = R+* x R and V = R-* x R do not cover the set ℝ² \ ℚ².
I'll give a proof that it is path connected

First notice that for any irrational p∈R, the horizontal line y=p and the vertical line x=p are paths that lie in our set ℝ² \ ℚ².
Then suppose we have the points (a,b), (c,d) ∈ℝ² \ ℚ² with (a,b) ≠ (c,d). Then at least one of a or b is irrational. WLOG suppose a is irrational.

Case 1: c is irrational
There exists f₁ : [0,1] ->ℝ² \ ℚ² a path from (a,b) to (a,b') where b' is irrational
f₂: [0,1] ->ℝ² \ ℚ² a path from (a,b') to (c,b')
f₃ : [0,1] ->ℝ² \ ℚ² a path from (c,b') to (c,d)
Thus define f(t) = { f₁(3t) when 0≤t≤1/3
{ f₂(3t-1) when 1/3<t≤2/3
{ f₃(3t-2) when 2/3<t≤1
So f(t) is a path that lies in ℝ² \ ℚ² from (a,b) to (c,d)

Case 2: d is irrational
There exists f₁ : [0,1] ->ℝ² \ ℚ² a path from (a,b) to (a,d)
f₂: [0,1] ->ℝ² \ ℚ² a path from (a,d) to (c,d)
Thus define f(t) = { f₁(2t) when 0≤t≤1/2
{ f₂(2t-1) when 1/2<t≤1
So f(t) is a path that lies in ℝ² \ ℚ² from (a,b) to (c,d)

Thus we conclude that ℝ² \ ℚ² is path-connected and hence connected

cdexswzaq is right !

I've read R² - ( RxQ ∪ QxR) instead of R² - Q².

My apologies...