From the documentation.

RegionPlot initially evaluates pred at a grid of equally spaced sample points specified by PlotPoints. Then it uses an adaptive algorithm to subdivide at most MaxRecursion times, attempting to find the boundaries of all regions in which pred is True.

and

You should realize that since it uses only a finite number of sample points, it is possible for RegionPlot to miss regions in which pred is True. To check your results, you should try increasing the settings for PlotPoints and MaxRecursion.

It works for x == y and the given ranges because the sample points include that line. If the range is changed to an irrational then RegionPlot is empty

RegionPlot[x == y, {x, -Sqrt[2], Sqrt[2]}, {y, -3/2, 3/2}]

I'm not sure about the best way, but these are methods I would employ:

Epilog:

solutions = Solve[{x^2 + y^2 == 1, x == y}, {x, y}];
ParametricPlot[ {{Cos[t], Sin[t]}, {t, t}}, {t, -Pi, Pi}, 
Epilog -> {Red, PointSize[0.05], Point[{x, y} /. solutions]}]

Show:

plot = ParametricPlot[ {{Cos[t], Sin[t]}, {t, t}}, {t, -Pi, Pi}]
points = Graphics[{Red, PointSize[0.05], Point[{x, y} /. solutions]}]
Show[plot, points] 

Hi Jay,

Can you give an example of

Sets defined by more complicated systems of equations -- the ones that need visualization -- are often difficult to parameterize.

Here is my workflow for your last example. It is not the best workflow, but how I approached it. Visualize the first two equations:

planes = 
 ContourPlot3D[{x + y + z == 1, x - 2 y - z == 3}, {x, -2, 2}, {y, -2,
    2}, {z, -2, 2}]

Look for a location where I might visualize the third equation:

FindInstance[(y - x^2)^2 + (z - x^3)^2 == 0, {x, y, z}, 6] // N

See if I can find conditions that there are real solutions to the last equation:

Reduce[(y - x^2)^2 + (z - x^3)^2 == 0, {x, y, z}, Reals]

Continue to see if there might be any real solutions other than 0:

Reduce[-x^4 + 2 x^2 y - y^2 > 0, {x, y}, Reals]

I conclude there is no intersection between the three surfaces.

For your second example:

eq1 = x y - z^2 - x;
eq2 = x y z - 2;
ineq3 = x^2 + y^2  + z^2 <= 5;

Sanity check to see if there are any intersections:

FindInstance[{eq1 == 0, eq2 == 0, ineq3}, {x, y, z}, Reals, 6] // N

Reduce[{eq1 == 0, eq2 == 0, ineq3}, {x, y, z}, Reals] // N

It looks like there is a narrow band around x=-1 that may have solutions in it.

Plot the first two surfaces in the region of interest:

surfaces = 
 ContourPlot3D[{eq1 == 0, eq2 == 0}, {x, -1.5, -1}, {y, -10, 
   10}, {z, -10, 10}]

plot the inequality region:

region = 
 RegionPlot3D[ineq3, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
  PlotStyle -> {Opacity[0.5], Orange}] (*I added Opacity on a second pass, not on the first work flow*)

VIsualize the intersection:

Show[surfaces, region]

Look for the curve. Copying and pasting as this is a one-time-only:

pp = ParametricPlot3D[{x, Root[-4 - x^3 #1^2 + x^3 #1^3 &, 1], 
   2/(x Root[-4 - x^3 #1^2 + x^3 #1^3 &, 1])}, {x, -1.15, -1.06}, 
  PlotStyle -> {Red, Thickness[0.02]}]

Visualize:

Show[surfaces, region, pp]

Thanks, W. Craig Carter. This is the best idea I've seen. I'll come back to this thread when I've had a chance to apply it more broadly. It will take me some time to see how well it generalizes for me. I'm still Mathematica challenged and mathematics challenged.

Thank you Gianluca Gorni. Your approach made fast work of all but one of the examples I posted. See notebook at the end. I'm always amazed at the level of help I can get from this community. I'm sorry I couldn't give you more than one thumbs up. Your answer deserves more.

Region[] and ImplicitRegion[] are amazing. You don't even need to tell them how many dimensions to return. I spent many hours pouring through the Visualization & Graphics section of the documentation. I should have been looking in the Geometry section. I posted an earlier question just asking what function(s) were most direct, but the moderators pulled that question because it didn't have any specific code questions.

I'm not sure what the conditionals are telling me in some of the results. I'll keep working on that. As for the system of equations that failed to render, Reduce[] gives an answer, but I'll have to figure out how to put it in a form ImplicitRegion[] can handle, if that's possible.

You could use Style

Show[
    Region[Style[ImplicitRegion[Subscript[\[ScriptCapitalR], 1], {x, y}], Red, PointSize@Large]],
    Region[Style[ImplicitRegion[Subscript[\[ScriptCapitalR], 2], {x, y}], Red, Thick]],
    PlotRange -> rng1, Axes -> True
 ]

Hi Jay,

You may be interested in this article from the Mathematica Journal. The article is based on this book by the same author. There is a link to download the notebooks used in the book.

Jay,

your basic question was perfectly answered by Hans Milton: Style in combination with Region - this is great (+1)!

Here I just have a remark:

When I started to try understanding Mathematica, (well, I am still trying!) I found this discussion on SE extremely useful. There you also will find some background regarding your question:

Also, should I care that Subscript[\[ScriptCapitalR], 1] doesn't turn black when defined?

The reason is that \[ScriptCapitalR] is not defined by this - and stays blue. What is done here instead is adding some additional definitions to Subscript.

So, Subscript[\[ScriptCapitalR], 1] is not a symbol, OwnValues[Subscript[\[ScriptCapitalR], 1]] does not work! The general advice is, not to use Subscript at all, except maybe at the very end of a calculation for displaying reasons.

Am I doing something wrong here? I think the two plots should be identical. If I'm not doing anything wrong, is there a workaround?

Using Simplify on the region condition is also a fix

reg2 = (y - x^2)^2 == 0 && (z - x^3)^2 == 0;

Region[
    Style[ImplicitRegion[reg2 // Simplify, {x, y, z}], Red, Thick],
    PlotRange -> 5, Axes -> True, AxesOrigin -> {0, 0, 0}
 ]

Why doesn't this work? The points {0,0,1} and {0,0,-1} are missing. I think I've tried all tricks in this thread that solved earlier problems work. So I may need a new trick.

Either half of the disjunction by itself renders as expected. The whole disjunction will render as expected If I delete one of the x or y equalities in the three-set conjunction. Adding parentheses to make the three-set conjunction into two two-set conjunctions doesn't work either. Nor does removing the parentheses and relying on order of operation.

I know there are several other ways to display the correct set. But I'm trying to find a easy general way to visualize varieties. And ImplicitRegion is a perfect solution. I just need to know how make it work as expected.

Thanks, Hans Milton. That's a good way to display a graphic I already know. But my goal is to find a generalized way for Mathematica to help me quickly understand sets defined by polynomial equations that would be hard to visualize.

You expressed my experience, perfectly, Gianluca. Thanks. It's awsome when it works and would be even more awsome if you could count on it to work. I still wonder if there is some hidden option or setting I could use every time that would always find the red dots in examples like the one I posted.