If the Na+ voltage gated channels remain open instead of getting deactivated during the re-polarization period, would the membrane potential become 0 since the Na+ ions would be constantly bringing into the cell while K+ is transported out of the cell, which makes them cancel each other out in a sense? However, I feel like there is something wrong with my reasoning but I don't know which part.
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" If the Na+ voltage gated channels remain open instead of getting deactivated during the re-polarization period, would the membrane potential become 0 ? "
No, if Na+ channels remain open then the repolarization period would never happen.
Mathematically we can prove the above with the Goldman–Hodgkin–Katz voltage equation
where 
approximate relative permeability values at the peak of a typical neuronal action potential are pK : pNa : pCl = 1 : 12 : 0.45 calculator
and thus we can see that the membrane potencial is much higher than the opposite for example
For a typical neuron at rest, pK : pNa : pCl = 1 : 0.05 : 0.45 calculator
and the only variable that we have changed was the relative membrane permeability for Na+
Hypothetically speaking if [K+]i = [K+]o = [Na+]i = [Na+]o = [Cl-]i = [Cl-]o the membrane potential in that case in fact would be zero.
However that is impossible because at rest, Na⁺/K⁺ ATPase constantly moves 3 Na+ ions out and moves 2 K+ ions in, and thus these gradients are maintain, so they cannot ever be the same.
Regarding your other question
What would happen to the membrane potential if a cell didn't have developed relative refractory period?
In that case the probability of repolarization and hiperpolarization to occur is diminished and thus it's possible to assume an extreme scenario of convulsions with death associated.
Example: Strychnine is a neurotoxin which acts as an antagonist of glycine and acetylcholine receptors
on the other hand, an excessive relative refractory period scenario, would result in a difficulty to achieve the threshold and depolarization
Example: mechanism of action of local anesthetics
by applying the principle above it is possible to reversibly decrease the rate of depolarization and repolarization of excitable membranes
As a side note (or curiosity), based upon that equation we can also modify Cl- values, and in that case we can speak about the mechanism of action of hypnotics and sedatives (such as benzodiazepines)
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$\begingroup$ Thank you so much of your detailed answer, I really appreciate it! If you don't mind, could you tell me why it's illogical to think that the membrane potential would become 0? I get that mathematically it changing the permeability of Na+ causes the membrane to stay depolarized but I still don't know why hypothetically speaking, in an environment where only Na+ and K+ are present inside and outside a membrane, the membrane potential wouldn't just be 0 when both voltage gated K+ and Na+ channels remain open and the total concentration of K+(inside+outside) = concentration of Na+(inside+outside). $\endgroup$ Dec 12, 2021 at 19:31
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$\begingroup$ if and only if, at rest, the Na+/K+ pump failed, then it would be possible such scenario.. but it has nothing to do with ion channels, because ion channels only open and close.. if there is no gradient then ions dont move. Sorry in advance, I think that i got a little off topic with my answer.. I've also edited it $\endgroup$– programDec 12, 2021 at 20:38
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$\begingroup$ Thank you..I gained a better understanding now! Just to make sure, is it right to think that if both Na+ & K+ voltage gated channels just remain open, depolarization wouldn't occur, and the membrane potential would decrease (compared to ~ +45) but it wouldn't become 0, because there is always a gradient. In this hypothetical situation, Na+ would keep coming in and K+ keeps going out, but the membrane potential doesn't depend on the opening/closing of voltage-gated channels, but rather the concentration gradient of different ions. Thank you again for answering my poorly worded question $\endgroup$ Dec 12, 2021 at 22:29