Lets say we have a glass vessel initially open to atmosphere.I can understand that the pressure inside will be equal to atmospheric pressure.,But now the vessel is closed. Now will the pressure inside exerted by the little air present equal the atmospheric pressure? If yes then how? How can little air column exert such a large air pressure?
The trapped air is compressed - it was squashed in there by air pressure in the first place. When you closed the lid, you kept it in its compressed state.
Think about those films you see of them squeezing people into the Tokyo Underground carriages. When the doors slide closed, the people are still squashed, aren't they?
Air pressure is the force that the air exerts on the container per unit area. You have to think about how this force on the container is being exerted. This force is being exerted through the air molecules bouncing off of the container walls, both inside and out. The pressure on the walls of the container is then going to depend on the rate at which the air molecules collide with the container wall and how much energy the have (in a simplified view of things). The amount air that is trapped in the container is certainly much less than the amount of air outside the container, but it can bounce between the walls quickly since the total volume on the inside is much less. Also, the amount of energy that they have is the same as that of the air outside in equilibrium (also I'm assuming that you didn't actively remove any of the air, so the air density is the same as well). So, they will be able to bounce off of the container walls at the same rate, exhibiting the same force per unit area as the air outside. Hence, the pressures are equal.