Why Is R1 Ignored in Thevenin Equivalent Calculation With Open Circuit Between A and B?

I'm a hobbyist and my electrical engineering knowledge couldn't fill a thimble, but I'm trying to understand something.
In the wikipedia article for Thevenin's theorem, there's an example calculation based on a circuit in which there are 4 resistors.  
There is a notation that reads:"(notice that R1 is not taken into consideration, as above calculations are done in an open-circuit condition between A and B, therefore no current flows through this part, which means there is no current through R1 and therefore no voltage drop along this part)"

There's evidently a huge gap in my understanding because I simply don't understand how these circuits can be considered equivalent if R1 is ignored.   As soon as the original circuit is closed, won't R1 have an effect on voltage and therefore behave differently than the Thevenin equivalent pictured?

You have jumped the gun a bit, the 1K does figure in the equivalent circuit, but only when computing the final equivalent resistive network. The first task in creating the Thevenin equivalent circuit is to find the equivalent voltage, and in this case R1 does not figure in computing that for the reason stated. If no current flows in a resistor, then the potential difference is zero, thus 7.5 volts appears at  point A with respect to B, which is the Thevenin equivalent voltage of the given circuit.Once the equivalent voltage is found, the voltage source is shorted, and the equivalent resistive network is computed, which does include R1.Once BOTH the equivalent voltage and resistance is found, then  you have a complete model. Now,  if you place a resistor across AB, current will flow through R1, and it's impact becomes evident.Cheers,
Richard

Richard's answer is very good. Another way to look at this is to do the Thevenin equivalent circuit after moving the point A to A' on the other side of R1. If you follow the steps correctly you should get the same 7.5V but you should have 1K resistance.Then you redraw the circuit with the Thevenin equivalent that you just found plus add R1 back in. then do the Thevenin equivalent of that circuit (which is trivial) and you should get the same result as before.This method of breaking the circuit down can be used to analyze very complex circuits.

Samson...I did this in a course recently and it took me a bit of time to get my head around, so I'll stick my two cents worth in too, even though Richard and Elizabeth have given very good answers.The Thevenin equivalent circuit reduces any combination of resistors and voltage sources down to ONE voltage Source and ONE resistance in series.  Looking at your diagrams - 1 is the original circuit.  First, assume that your points A and B are totally open circuit.  So no current will flow in R1.  The voltage of 15V is across a 2K and two 1K resistors, which will divide the voltage by 2 to 7.5V as shown in diagram 2.  This is the MAXIMUM voltage you will get across AB in any circumstances.  (If you constructed this circuit and measured AB with a digital multimeter, you'd get as near as dammit to this figure, as a DMM has a resistance of 10 Megohms, which is high enough not to matter much in a circuit like this).Now you need to know the effective resistance that is in series with  the voltage source, and not R1 comes into play.  the technique is to replace the voltage source with a short circuit as in diagram 3.  Imagine that you put your DMM on ohms and measure between A and B then.  You will see 1K (R1), plus 2K (R4) and 2K (the two 1K resistors (R2 + R3)) in parallel.  2K in parallel with 2K is 1K.  so you have 1K (R1) plus 1K (the R2/3/4 combination) in series, which makes 2K.  So the Thevenin equivalent for this circuit is a 7.5V source in series with a 2K resistance (diagram 4).  You can use Kirchhoff's laws to work this out another way, but Thevenin's theorem is an easier way of doing this.As an aside, if you're doing Thevenin's you will soon encounter Nortons Theorem as well.  Norton's theorem will reduce any circuit down to a current source in paralell with a resistor.  Once you have the Thevenin equivalent it's easy.  What current will flow if you short A and B?  7.5V / 2Kohms = 3.75 mA.  What voltage is across A and B with nothing accross them? 7.5V.  So the Norton resistor will be the resistance that gets 7.5 Volts across it with 3.75mA flowing through it.  R = V/I = 7.5 / 0.00375 = 2Kohms.  The resistances are the same, you just have to work out the current (or, if you're working form Norton > Thevenin, the voltage).  This is covered in the wikipedia article as well.Hope this helps, yell out if anything is not clear...     //  David

Thanks so much for your replies. It took a while but it finally sunk in.  I'm so used to measuring everything, I confused measurement with applying a formula.  Sort of like actually measuring area of a triangle instead of applying the Pythagorean theorem.
After reading your comments and stepping through the calculation half a dozen times, it finally dawned on me. Actually seeing the process of solving for Norton was what finally helped me back up and see why R1 is initially ignored.
The irony here is that the circuit I need to build an equivalent for is not well documented (and it's obscured so I have to resort to measurement anyway. Lol.  
Again, thanks for the thoughtful detailed responses!

Samson, thanks for YOUR reply.  A lot of the time this is a thankless task, we put a reply out there and never get any reply as to whether the questioner has seen it, let alone whether it helped them.  To get a reply and thanks like yours makes it worthwhile for us.  Feel free to ask us anything anytime!
Oh and BTW I see in my 4th paragraph, I put "and NOT R1 comes into play..." instead of "and NOW R1...Sorry for that, hope you saw past my typo....Cheers // David

David,
In my particular field, I'm usually on the other end of this exchange, so I totally understand what you mean.  And when I originally ready your comment my brain must have automatically substituted "not" for "now" because I got your meaning and didn't even notice the typo.Cheers,-Samson-