A battery is DC, and a transformer requires magnetic flux to work.
But assuming you mean the input voltage was from a 5 V AC source, and the turns ratio was 2000:1, then you can, sort of, but how much current can be provided is limited by core saturation. Basically, current flows in the secondary winding because the magnetic field in the iron core changes. That change is due to small regions of the core realigning in conjunction with the applied voltage (there's sort of a cascade effect throughout the core, but it's very complicated). Once all the regions are aligned in one direction, that's it... voltage drops to 0 and no more current can flow. So for an applied current that's greater than the transformer's rating, you get a rise in voltage toward 10000 V, but then the core saturates, and it drops to 0 before ever getting there. Then the waveform goes the other way, you get a rise, but then it saturates in that direction....
Now, you might think you could overcome this by simply making a bigger core, but then there are losses due to things like hysteresis which reduce the efficiency of the transformer. And because the secondary winding is inductive, there is a dependence between the current and voltage. If you have a capacitive load of the right size, however, you can cancel this out and get a dangerous condition called ferroresonance, in which the voltage is almost uncapped... and then you can get your 10000 V. It's a serious issue on the power system, and it's why you never de-energize a voltage regulator and a feeder cable at the same time on single phase switches. (On 3 phase, it's not usually so bad, because there isn't enough time to build up a big overvoltage and damage anything.) If you're designing substations, please keep this in mind.
