If the eye pattern is open, then ISI might be present but does not distort the received signal. Let us say that "symbols might be separated" and we say that there is no ISI (in fact, there is low ISI and it does not create errors). If the eye pattern is closed, that means that ISI is strong enough to "mix" symbols: the influence of adjacent symbols+channel modifies the value of the received symbol so much that it crosses the threshold and is detected incorrectly (eg: in a +1/-1 constellation, +1 suffers distortion enough to become, for example, -0.2, and hence it is detected as -1; if the eye pattern is open, the +1 value never gets below 0 and never is detected as -1). Notice that this refers to the effect of the channel, not the noise. The noise might create any distortion...

Hope it helps.

Dear Researcher

please refer to this attachments if it is helpful for you.

Regards

http://www.eng.auburn.edu/~troppel/courses/3060%202009A%20Wireless%20Lab%20Spring%2009/Manual_Exp7-eye-patterns.pdf

https://en.wikipedia.org/wiki/Eye_pattern

Adding to what has been said by the colleagues, Assume that you transmit a sequence of ones and zeros using bipolar nonreturn zero line coding. Which means that the logic 1 is represented by rectangular positive pulse and the zero is represented by a negative rectangular pulse. Displaying these two pulses on the screen of the oscilloscope with time base of bit time for every symbol results in an rectangular eye pattern. This is the most ideal eye pattern. Because of symbol distortion and contamination by the noise the rectangular pulse appears more or less as a half sine wave with zero transitions not lying on the edges of the oscilloscope and amplitude less than of the original rectangular eye. So, there is amplitude and phase distortion.

In order to decode the distorted symbols into logic value one chooses the most appropriate points of the symbols which is the points tat have the maximum discrimination distance. These points lie on the center line of the eye where the eye opening is largest. 

The intersymbol interference ISI is deviation of the center point of the symbol from their ideal value. That is when the eye opening decreases the ISI increases and vice verse,

If the value of symbol at the center line of the eye is greater than the decision value,   threshold, which in this example is zero value of the signal,it will be coined as logic one and if it is lower than the threshold it be coined as zero. The distortion and noise can be so high that a logic one symbol resides under the decision value. In this case there will be a false decision on the bit value. This also holds for the symbol representing zero.

One of the most causes of ISI is the limited band width of the transmission system.

It is so that at least the bandwidth must be greater than half of the bit rate. If this condition is not satisfied, the ISI will be large and the bit error rate will increase because of the false decision on the symbols of the eye.Best wishes

I agree with the answers of the previous colleagues.

Eye diagram is a measure of the distortion of the signal. You start with an ideal rectangular pulse and then distortion of the signal due to channel effects (e.g. ISI, channel dispersion etc.) causes distortion in the ideal rectangular signal causing it to change shape from rectangle to elliptical eye shape. The opening of the eye tells you whether you can still distinguish between 1 and 0 based on threshold detection mechanism. 

When you observe a bad eye diagram, how do you differentiate ISI from unmatched channel (noise due to reflections)? What can be done to reduce ISI?